x^2-45x+400=0

Solution for x^2-45x+400=0 equation:

x^2-45x+400=0

a = 1; b = -45; c = +400;
Δ = b2-4ac
Δ = -452-4·1·400
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-5\sqrt{17}}{2*1}=\frac{45-5\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+5\sqrt{17}}{2*1}=\frac{45+5\sqrt{17}}{2}
$