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(x)(x+12)=160
We move all terms to the left:
(x)(x+12)-(160)=0
We multiply parentheses
x^2+12x-160=0
a = 1; b = 12; c = -160;
Δ = b2-4ac
Δ = 122-4·1·(-160)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*1}=\frac{-40}{2} =-20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*1}=\frac{16}{2} =8 $
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