-3x^2+3x+10=0

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Solution for -3x^2+3x+10=0 equation: 



-3x^2+3x+10=0

a = -3; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-3)·10
Δ = 129
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{-(3)-\sqrt{129}}{2*-3}=\frac{-3-\sqrt{129}}{-6}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{129}}{2*-3}=\frac{-3+\sqrt{129}}{-6}
$

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