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1344=(2x-6)(x+8)
We move all terms to the left:
1344-((2x-6)(x+8))=0
We multiply parentheses ..
-((+2x^2+16x-6x-48))+1344=0
We calculate terms in parentheses: -((+2x^2+16x-6x-48)), so:We get rid of parentheses
(+2x^2+16x-6x-48)
We get rid of parentheses
2x^2+16x-6x-48
We add all the numbers together, and all the variables
2x^2+10x-48
Back to the equation:
-(2x^2+10x-48)
-2x^2-10x+48+1344=0
We add all the numbers together, and all the variables
-2x^2-10x+1392=0
a = -2; b = -10; c = +1392;
Δ = b2-4ac
Δ = -102-4·(-2)·1392
Δ = 11236
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{11236}=106$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-106}{2*-2}=\frac{-96}{-4} =+24 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+106}{2*-2}=\frac{116}{-4} =-29 $
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