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10=14x^2
We move all terms to the left:
10-(14x^2)=0
a = -14; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-14)·10
Δ = 560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{35}}{2*-14}=\frac{0-4\sqrt{35}}{-28} =-\frac{4\sqrt{35}}{-28} =-\frac{\sqrt{35}}{-7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{35}}{2*-14}=\frac{0+4\sqrt{35}}{-28} =\frac{4\sqrt{35}}{-28} =\frac{\sqrt{35}}{-7} $
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