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14x^2+10x=10
We move all terms to the left:
14x^2+10x-(10)=0
a = 14; b = 10; c = -10;
Δ = b2-4ac
Δ = 102-4·14·(-10)
Δ = 660
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{660}=\sqrt{4*165}=\sqrt{4}*\sqrt{165}=2\sqrt{165}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{165}}{2*14}=\frac{-10-2\sqrt{165}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{165}}{2*14}=\frac{-10+2\sqrt{165}}{28} $
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