4m^2=-10m+19

Solution for 4m^2=-10m+19 equation:

4m^2=-10m+19

We move all terms to the left:

4m^2-(-10m+19)=0

We get rid of parentheses

4m^2+10m-19=0

a = 4; b = 10; c = -19;
Δ = b2-4ac
Δ = 102-4·4·(-19)
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{101}}{2*4}=\frac{-10-2\sqrt{101}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{101}}{2*4}=\frac{-10+2\sqrt{101}}{8}
$