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2x^2+240=88x
We move all terms to the left:
2x^2+240-(88x)=0
a = 2; b = -88; c = +240;
Δ = b2-4ac
Δ = -882-4·2·240
Δ = 5824
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5824}=\sqrt{64*91}=\sqrt{64}*\sqrt{91}=8\sqrt{91}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-88)-8\sqrt{91}}{2*2}=\frac{88-8\sqrt{91}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-88)+8\sqrt{91}}{2*2}=\frac{88+8\sqrt{91}}{4} $
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