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3x^2+42x+145=0
a = 3; b = 42; c = +145;
Δ = b2-4ac
Δ = 422-4·3·145
Δ = 24
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{6}}{2*3}=\frac{-42-2\sqrt{6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{6}}{2*3}=\frac{-42+2\sqrt{6}}{6} $
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