40x+4x^2=156

Solution for 40x+4x^2=156 equation:

40x+4x^2=156

We move all terms to the left:

40x+4x^2-(156)=0

a = 4; b = 40; c = -156;
Δ = b2-4ac
Δ = 402-4·4·(-156)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-64}{2*4}=\frac{-104}{8}
=-13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+64}{2*4}=\frac{24}{8}
=3
$