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142=2x^2
We move all terms to the left:
142-(2x^2)=0
a = -2; b = 0; c = +142;
Δ = b2-4ac
Δ = 02-4·(-2)·142
Δ = 1136
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{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{71}}{2*-2}=\frac{0-4\sqrt{71}}{-4} =-\frac{4\sqrt{71}}{-4} =-\frac{\sqrt{71}}{-1}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{71}}{2*-2}=\frac{0+4\sqrt{71}}{-4} =\frac{4\sqrt{71}}{-4} =\frac{\sqrt{71}}{-1}
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