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(x+5)(2x-4)=120
We move all terms to the left:
(x+5)(2x-4)-(120)=0
We multiply parentheses ..
(+2x^2-4x+10x-20)-120=0
We get rid of parentheses
2x^2-4x+10x-20-120=0
We add all the numbers together, and all the variables
2x^2+6x-140=0
a = 2; b = 6; c = -140;
Δ = b2-4ac
Δ = 62-4·2·(-140)
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-34}{2*2}=\frac{-40}{4} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+34}{2*2}=\frac{28}{4} =7 $
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