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-16x^2+112x=160
We move all terms to the left:
-16x^2+112x-(160)=0
a = -16; b = 112; c = -160;
Δ = b2-4ac
Δ = 1122-4·(-16)·(-160)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-48}{2*-16}=\frac{-160}{-32} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+48}{2*-16}=\frac{-64}{-32} =+2 $
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