3m^2+15m+21=4

Solution for 3m^2+15m+21=4 equation:

3m^2+15m+21=4

We move all terms to the left:

3m^2+15m+21-(4)=0

We add all the numbers together, and all the variables

3m^2+15m+17=0

a = 3; b = 15; c = +17;
Δ = b2-4ac
Δ = 152-4·3·17
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{21}}{2*3}=\frac{-15-\sqrt{21}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{21}}{2*3}=\frac{-15+\sqrt{21}}{6}
$