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4x^2+40x-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 $
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