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(x+4)(2x+5)=128
We move all terms to the left:
(x+4)(2x+5)-(128)=0
We multiply parentheses ..
(+2x^2+5x+8x+20)-128=0
We get rid of parentheses
2x^2+5x+8x+20-128=0
We add all the numbers together, and all the variables
2x^2+13x-108=0
a = 2; b = 13; c = -108;
Δ = b2-4ac
Δ = 132-4·2·(-108)
Δ = 1033
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{1033}}{2*2}=\frac{-13-\sqrt{1033}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{1033}}{2*2}=\frac{-13+\sqrt{1033}}{4} $
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