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.x(2x-8)+10(2x-8)=0
We multiply parentheses
2x^2-8x+20x-80=0
We add all the numbers together, and all the variables
2x^2+12x-80=0
a = 2; b = 12; c = -80;
Δ = b2-4ac
Δ = 122-4·2·(-80)
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*2}=\frac{-40}{4} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*2}=\frac{16}{4} =4 $
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