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160=x(x-12)
We move all terms to the left:
160-(x(x-12))=0
We calculate terms in parentheses: -(x(x-12)), so:We get rid of parentheses
x(x-12)
We multiply parentheses
x^2-12x
Back to the equation:
-(x^2-12x)
-x^2+12x+160=0
We add all the numbers together, and all the variables
-1x^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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