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3x^2+34x-45=0
a = 3; b = 34; c = -45;
Δ = b2-4ac
Δ = 342-4·3·(-45)
Δ = 1696
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-4\sqrt{106}}{2*3}=\frac{-34-4\sqrt{106}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+4\sqrt{106}}{2*3}=\frac{-34+4\sqrt{106}}{6} $
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