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(2x-3)(x+40)=0
We multiply parentheses ..
(+2x^2+80x-3x-120)=0
We get rid of parentheses
2x^2+80x-3x-120=0
We add all the numbers together, and all the variables
2x^2+77x-120=0
a = 2; b = 77; c = -120;
Δ = b2-4ac
Δ = 772-4·2·(-120)
Δ = 6889
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{6889}=83$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(77)-83}{2*2}=\frac{-160}{4} =-40 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(77)+83}{2*2}=\frac{6}{4} =1+1/2 $
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