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2x^2+8x=10
We move all terms to the left:
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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