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(8x)(2x+40)=180
We move all terms to the left:
(8x)(2x+40)-(180)=0
We multiply parentheses
16x^2+320x-180=0
a = 16; b = 320; c = -180;
Δ = b2-4ac
Δ = 3202-4·16·(-180)
Δ = 113920
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{113920}=\sqrt{256*445}=\sqrt{256}*\sqrt{445}=16\sqrt{445}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-16\sqrt{445}}{2*16}=\frac{-320-16\sqrt{445}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+16\sqrt{445}}{2*16}=\frac{-320+16\sqrt{445}}{32} $
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