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14x^2-12x=35
We move all terms to the left:
14x^2-12x-(35)=0
a = 14; b = -12; c = -35;
Δ = b2-4ac
Δ = -122-4·14·(-35)
Δ = 2104
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2104}=\sqrt{4*526}=\sqrt{4}*\sqrt{526}=2\sqrt{526}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{526}}{2*14}=\frac{12-2\sqrt{526}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{526}}{2*14}=\frac{12+2\sqrt{526}}{28} $
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