9x^2+42x-95=0

Solution for 9x^2+42x-95=0 equation:

9x^2+42x-95=0

a = 9; b = 42; c = -95;
Δ = b2-4ac
Δ = 422-4·9·(-95)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-72}{2*9}=\frac{-114}{18}
=-6+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+72}{2*9}=\frac{30}{18}
=1+2/3
$