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3x^2+26x+32=0
a = 3; b = 26; c = +32;
Δ = b2-4ac
Δ = 262-4·3·32
Δ = 292
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{73}}{2*3}=\frac{-26-2\sqrt{73}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{73}}{2*3}=\frac{-26+2\sqrt{73}}{6} $
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