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(1008)=(2x-4)(x+8)
We move all terms to the left:
(1008)-((2x-4)(x+8))=0
We multiply parentheses ..
-((+2x^2+16x-4x-32))+1008=0
We calculate terms in parentheses: -((+2x^2+16x-4x-32)), so:We get rid of parentheses
(+2x^2+16x-4x-32)
We get rid of parentheses
2x^2+16x-4x-32
We add all the numbers together, and all the variables
2x^2+12x-32
Back to the equation:
-(2x^2+12x-32)
-2x^2-12x+32+1008=0
We add all the numbers together, and all the variables
-2x^2-12x+1040=0
a = -2; b = -12; c = +1040;
Δ = b2-4ac
Δ = -122-4·(-2)·1040
Δ = 8464
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{8464}=92$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-92}{2*-2}=\frac{-80}{-4} =+20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+92}{2*-2}=\frac{104}{-4} =-26 $
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