G(x)=4(x-5)+x2

Solution for G(x)=4(x-5)+x2 equation:

(G)=4(G-5)+G2

We move all terms to the left:

(G)-(4(G-5)+G2)=0


We calculate terms in parentheses: -(4(G-5)+G2), so:

4(G-5)+G2

We add all the numbers together, and all the variables

G^2+4(G-5)

We multiply parentheses

G^2+4G-20

Back to the equation:

-(G^2+4G-20)


We get rid of parentheses

-G^2+G-4G+20=0

We add all the numbers together, and all the variables

-1G^2-3G+20=0

a = -1; b = -3; c = +20;
Δ = b2-4ac
Δ = -32-4·(-1)·20
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{89}}{2*-1}=\frac{3-\sqrt{89}}{-2}
$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*-1}=\frac{3+\sqrt{89}}{-2}
$