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4x^2+40x+64=0
a = 4; b = 40; c = +64;
Δ = b2-4ac
Δ = 402-4·4·64
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-24}{2*4}=\frac{-64}{8} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+24}{2*4}=\frac{-16}{8} =-2 $
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