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9x^2+6x=13
We move all terms to the left:
9x^2+6x-(13)=0
a = 9; b = 6; c = -13;
Δ = b2-4ac
Δ = 62-4·9·(-13)
Δ = 504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{14}}{2*9}=\frac{-6-6\sqrt{14}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{14}}{2*9}=\frac{-6+6\sqrt{14}}{18} $
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