3m2-22m+25=0

Solution for 3m2-22m+25=0 equation:

3m^2-22m+25=0

a = 3; b = -22; c = +25;
Δ = b2-4ac
Δ = -222-4·3·25
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{46}}{2*3}=\frac{22-2\sqrt{46}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{46}}{2*3}=\frac{22+2\sqrt{46}}{6}
$