5m^2-42m+15=0

Solution for 5m^2-42m+15=0 equation:

5m^2-42m+15=0

a = 5; b = -42; c = +15;
Δ = b2-4ac
Δ = -422-4·5·15
Δ = 1464
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:

$\sqrt{\Delta}=\sqrt{1464}=\sqrt{4*366}=\sqrt{4}*\sqrt{366}=2\sqrt{366}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{366}}{2*5}=\frac{42-2\sqrt{366}}{10}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{366}}{2*5}=\frac{42+2\sqrt{366}}{10}
$