3x^2-90x+500=0

Solution for 3x^2-90x+500=0 equation:

3x^2-90x+500=0

a = 3; b = -90; c = +500;
Δ = b2-4ac
Δ = -902-4·3·500
Δ = 2100
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:

$\sqrt{\Delta}=\sqrt{2100}=\sqrt{100*21}=\sqrt{100}*\sqrt{21}=10\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-10\sqrt{21}}{2*3}=\frac{90-10\sqrt{21}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+10\sqrt{21}}{2*3}=\frac{90+10\sqrt{21}}{6}
$