4m^2+16m-9=0

Solution for 4m^2+16m-9=0 equation:

4m^2+16m-9=0

a = 4; b = 16; c = -9;
Δ = b2-4ac
Δ = 162-4·4·(-9)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*4}=\frac{-36}{8}
=-4+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*4}=\frac{4}{8}
=1/2
$