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2x^2-8=-6x
We move all terms to the left:
2x^2-8-(-6x)=0
We get rid of parentheses
2x^2+6x-8=0
a = 2; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·2·(-8)
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*2}=\frac{4}{4} =1 $
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