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2x^2+8x=216
We move all terms to the left:
2x^2+8x-(216)=0
a = 2; b = 8; c = -216;
Δ = b2-4ac
Δ = 82-4·2·(-216)
Δ = 1792
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16\sqrt{7}}{2*2}=\frac{-8-16\sqrt{7}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16\sqrt{7}}{2*2}=\frac{-8+16\sqrt{7}}{4} $
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