-m^2+42m-30=0

Solution for -m^2+42m-30=0 equation:

-m^2+42m-30=0

We add all the numbers together, and all the variables

-1m^2+42m-30=0

a = -1; b = 42; c = -30;
Δ = b2-4ac
Δ = 422-4·(-1)·(-30)
Δ = 1644
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:

$\sqrt{\Delta}=\sqrt{1644}=\sqrt{4*411}=\sqrt{4}*\sqrt{411}=2\sqrt{411}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{411}}{2*-1}=\frac{-42-2\sqrt{411}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{411}}{2*-1}=\frac{-42+2\sqrt{411}}{-2}
$