2x^2+42x+216=0

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Solution for 2x^2+42x+216=0 equation: 



2x^2+42x+216=0

a = 2; b = 42; c = +216;
Δ = b2-4ac
Δ = 422-4·2·216
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6}{2*2}=\frac{-48}{4}
=-12
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6}{2*2}=\frac{-36}{4}
=-9
$

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