M(x)=3x^2+15x+4

Solution for M(x)=3x^2+15x+4 equation:

(M)=3M^2+15M+4

We move all terms to the left:

(M)-(3M^2+15M+4)=0

We get rid of parentheses

-3M^2+M-15M-4=0

We add all the numbers together, and all the variables

-3M^2-14M-4=0

a = -3; b = -14; c = -4;
Δ = b2-4ac
Δ = -142-4·(-3)·(-4)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{37}}{2*-3}=\frac{14-2\sqrt{37}}{-6}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{37}}{2*-3}=\frac{14+2\sqrt{37}}{-6}
$