9x^2-340x+3025=0

Solution for 9x^2-340x+3025=0 equation:

9x^2-340x+3025=0

a = 9; b = -340; c = +3025;
Δ = b2-4ac
Δ = -3402-4·9·3025
Δ = 6700
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:

$\sqrt{\Delta}=\sqrt{6700}=\sqrt{100*67}=\sqrt{100}*\sqrt{67}=10\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-340)-10\sqrt{67}}{2*9}=\frac{340-10\sqrt{67}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-340)+10\sqrt{67}}{2*9}=\frac{340+10\sqrt{67}}{18}
$