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160=(2x+4)(x)
We move all terms to the left:
160-((2x+4)(x))=0
We calculate terms in parentheses: -((2x+4)x), so:We get rid of parentheses
(2x+4)x
We multiply parentheses
2x^2+4x
Back to the equation:
-(2x^2+4x)
-2x^2-4x+160=0
a = -2; b = -4; c = +160;
Δ = b2-4ac
Δ = -42-4·(-2)·160
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-36}{2*-2}=\frac{-32}{-4} =+8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+36}{2*-2}=\frac{40}{-4} =-10 $
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