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(2x-4)(x-4)=96
We move all terms to the left:
(2x-4)(x-4)-(96)=0
We multiply parentheses ..
(+2x^2-8x-4x+16)-96=0
We get rid of parentheses
2x^2-8x-4x+16-96=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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-28}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+28}{2*2}=\frac{40}{4} =10 $
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