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6x^2+8x=13
We move all terms to the left:
6x^2+8x-(13)=0
a = 6; b = 8; c = -13;
Δ = b2-4ac
Δ = 82-4·6·(-13)
Δ = 376
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{94}}{2*6}=\frac{-8-2\sqrt{94}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{94}}{2*6}=\frac{-8+2\sqrt{94}}{12} $
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