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5x^2=62
We move all terms to the left:
5x^2-(62)=0
a = 5; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·5·(-62)
Δ = 1240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1240}=\sqrt{4*310}=\sqrt{4}*\sqrt{310}=2\sqrt{310}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{310}}{2*5}=\frac{0-2\sqrt{310}}{10} =-\frac{2\sqrt{310}}{10} =-\frac{\sqrt{310}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{310}}{2*5}=\frac{0+2\sqrt{310}}{10} =\frac{2\sqrt{310}}{10} =\frac{\sqrt{310}}{5} $
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