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2x^2+8x=72
We move all terms to the left:
2x^2+8x-(72)=0
a = 2; b = 8; c = -72;
Δ = b2-4ac
Δ = 82-4·2·(-72)
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{10}}{2*2}=\frac{-8-8\sqrt{10}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{10}}{2*2}=\frac{-8+8\sqrt{10}}{4} $
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