X^2+301x+6242.25=0

Solution for X^2+301x+6242.25=0 equation:

X^2+301X+6242.25=0

a = 1; b = 301; c = +6242.25;
Δ = b2-4ac
Δ = 3012-4·1·6242.25
Δ = 65632
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:

$\sqrt{\Delta}=\sqrt{65632}=\sqrt{16*4102}=\sqrt{16}*\sqrt{4102}=4\sqrt{4102}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(301)-4\sqrt{4102}}{2*1}=\frac{-301-4\sqrt{4102}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(301)+4\sqrt{4102}}{2*1}=\frac{-301+4\sqrt{4102}}{2}
$