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4x^2+40x+36=0
a = 4; b = 40; c = +36;
Δ = b2-4ac
Δ = 402-4·4·36
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-32}{2*4}=\frac{-72}{8} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+32}{2*4}=\frac{-8}{8} =-1 $
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