# From symbolic to numerical results for quadratic equation

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laura bagnale on 17 Jun 2021
Edited: Walter Roberson on 19 Jun 2021
Hello everyone,
I hope someone could help me, I'm starting using the symbolic toolbox but I have some difficulties.
I wrote this quadratic equation in symbolic form:
syms x y a1 b1 c1 d1 e1 f1
a = d1*y^2 + (e1 + b1*x)*y + a1*x^2 + c1*x + f1
then I tried to solve it:
ht=matlabFunction(a)
eqn = d1.*y.^2+(e1+b1.*x).*y+(a1.*x.^2+c1.*x+f1) == 0
S = solve(eqn, y)
I got what I expected from the formula
[-(e1 + b1*x - sqrt(b1^2*x^2 + 2*b1*e1*x + e1^2 - 4*a1*d1*x^2 - 4*c1*d1*x - 4*d1*f1))/(2*d1); -(e1 + b1*x + sqrt(b1^2*x^2 + 2*b1*e1*x + e1^2 - 4*a1*d1*x^2 - 4*c1*d1*x - 4*d1*f1))/(2*d1)]
However I don't know how to get the numerical real solutions.
I know the numerical values for the coeffcients a1, b1, c1, d1, e1, f1.
I tried to write the equation in this way:
syms x y
a = d1*y^2 + (e1 + b1*x)*y + a1*x^2 + c1*x + f1
ht=matlabFunction(a)
eqn = d1.*y.^2+(e1+b1.*x).*y+(a1.*x.^2+c1.*x+f1) == 0
S = solve(eqn, y, 'Real', true)
but the formula appears in the same form, with numbers in the places of the coefficients but with no solutions.
Thank you very much in advance.
Laura
##### 2 CommentsShowHide 1 older comment
laura bagnale on 17 Jun 2021
Thank you a lot KSSV for your quick reply! I will try with subs as you suggested.
Yes, I defined the coefficients.
This is the part of the code regarding my question, the previous one is very long. I obtained pcell and qcell by using poly22 to interpolate 5 points with a surface in space.
[p00, p10, p01, p20, p11, p02] = pcell{:}
[q00, q10, q01, q20, q11, q02] = qcell{:}
format short
a1 = p20.*q11.*2.0-p11.*q20.*2.0
b1 = p20.*q20.*4.0-p02.*q20.*4.0
c1 = p10.*q11-p01.*q20.*2.0-p11.*q10+p20.*q01.*2.0
d1 = p11.*q20.*2.0-p02.*q11.*2.0
e1 = p11.*q01-p02.*q10.*2.0-p01.*q11+p10.*q20.*2.0
f1 = p10.*q01-p01.*q10
a1 = -0.0870
b1 = -0.2519
c1 = -0.0847
d1 = -0.0495
e1 = -0.0386
f1 = -0.0081
g = @(x,y) a1.*x.^2 + b1.*x.*y + c1.*x + d1.*y.^2 + e1.*y + f1
syms x y a1 b1 c1 d1 e1 f1
eqn = a1.*x.^2 + b1.*x.*y + c1.*x + d1.*y.^2 + e1.*y + f1
a = collect(eqn, y)
ht=matlabFunction(a)
eqn = d1.*y.^2+(e1+b1.*x).*y+(a1.*x.^2+c1.*x+f1) == 0
S = solve(eqn, y)
syms x y a1 b1 c1 d1 e1 f1
eqn = a1.*x.^2 + b1.*x.*y + c1.*x + d1.*y.^2 + e1.*y + f1
a = collect(eqn, y)
ht=matlabFunction(a)
eqn = d1.*y.^2+(e1+b1.*x).*y+(a1.*x.^2+c1.*x+f1) == 0
S = solve(eqn, y)
Thank you very much for your help and support.
Laura

Stephan on 17 Jun 2021
Edited: Stephan on 17 Jun 2021
You could use symbolic functions:
syms y(a,b,x)
y(a,b,x) = a*x^2 - b*x
y(a, b, x) = % Calculate values for a=1, b=2 and x in a range from 1:4 with stepwide 0.5
result_symbolic = y(1,2,1:0.5:4)
result_symbolic = if you want that numbers (still symbolic) use double
result_numeric = double(result_symbolic)
result_numeric = 1×7
-1.0000 -0.7500 0 1.2500 3.0000 5.2500 8.0000
##### 2 CommentsShowHide 1 older comment
laura bagnale on 17 Jun 2021
I mean my difficulty is that the function is quadratic with 2 variables. I didn't find something similar from the MathWorks.
Thanks!

R2021a

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