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(x+10)(2x-4)=0
We multiply parentheses ..
(+2x^2-4x+20x-40)=0
We get rid of parentheses
2x^2-4x+20x-40=0
We add all the numbers together, and all the variables
2x^2+16x-40=0
a = 2; b = 16; c = -40;
Δ = b2-4ac
Δ = 162-4·2·(-40)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*2}=\frac{-40}{4} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*2}=\frac{8}{4} =2 $
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