m^2-10m+15=0

Solution for m^2-10m+15=0 equation:

m^2-10m+15=0

a = 1; b = -10; c = +15;
Δ = b2-4ac
Δ = -102-4·1·15
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{10}}{2*1}=\frac{10-2\sqrt{10}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{10}}{2*1}=\frac{10+2\sqrt{10}}{2}
$