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3x^2+2x-9=0
a = 3; b = 2; c = -9;
Δ = b2-4ac
Δ = 22-4·3·(-9)
Δ = 112
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{7}}{2*3}=\frac{-2-4\sqrt{7}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{7}}{2*3}=\frac{-2+4\sqrt{7}}{6} $
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