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10=2x^2-8x
We move all terms to the left:
10-(2x^2-8x)=0
We get rid of parentheses
-2x^2+8x+10=0
a = -2; b = 8; c = +10;
Δ = b2-4ac
Δ = 82-4·(-2)·10
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*-2}=\frac{-20}{-4} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*-2}=\frac{4}{-4} =-1 $
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