3m^2+16m-21=0

Solution for 3m^2+16m-21=0 equation:

3m^2+16m-21=0

a = 3; b = 16; c = -21;
Δ = b2-4ac
Δ = 162-4·3·(-21)
Δ = 508
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:

$\sqrt{\Delta}=\sqrt{508}=\sqrt{4*127}=\sqrt{4}*\sqrt{127}=2\sqrt{127}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{127}}{2*3}=\frac{-16-2\sqrt{127}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{127}}{2*3}=\frac{-16+2\sqrt{127}}{6}
$