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x^2+16x=132
We move all terms to the left:
x^2+16x-(132)=0
a = 1; b = 16; c = -132;
Δ = b2-4ac
Δ = 162-4·1·(-132)
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-28}{2*1}=\frac{-44}{2} =-22 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+28}{2*1}=\frac{12}{2} =6 $
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