3x^2+7x-38=0

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Solution for 3x^2+7x-38=0 equation: 



3x^2+7x-38=0

a = 3; b = 7; c = -38;
Δ = b2-4ac
Δ = 72-4·3·(-38)
Δ = 505
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{505}}{2*3}=\frac{-7-\sqrt{505}}{6}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{505}}{2*3}=\frac{-7+\sqrt{505}}{6}
$

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