5x^2-410x+300=0

Solution for 5x^2-410x+300=0 equation:

5x^2-410x+300=0

a = 5; b = -410; c = +300;
Δ = b2-4ac
Δ = -4102-4·5·300
Δ = 162100
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:

$\sqrt{\Delta}=\sqrt{162100}=\sqrt{100*1621}=\sqrt{100}*\sqrt{1621}=10\sqrt{1621}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-410)-10\sqrt{1621}}{2*5}=\frac{410-10\sqrt{1621}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-410)+10\sqrt{1621}}{2*5}=\frac{410+10\sqrt{1621}}{10}
$