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4x^2+56x+132=0.
a = 4; b = 56; c = +132;
Δ = b2-4ac
Δ = 562-4·4·132
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-32}{2*4}=\frac{-88}{8} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+32}{2*4}=\frac{-24}{8} =-3 $
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