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3x^2+26x+7=0
a = 3; b = 26; c = +7;
Δ = b2-4ac
Δ = 262-4·3·7
Δ = 592
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4\sqrt{37}}{2*3}=\frac{-26-4\sqrt{37}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4\sqrt{37}}{2*3}=\frac{-26+4\sqrt{37}}{6} $
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