3x^2+19x^2=141

Solution for 3x^2+19x^2=141 equation:

3x^2+19x^2=141

We move all terms to the left:

3x^2+19x^2-(141)=0

We add all the numbers together, and all the variables

22x^2-141=0

a = 22; b = 0; c = -141;
Δ = b2-4ac
Δ = 02-4·22·(-141)
Δ = 12408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12408}=\sqrt{4*3102}=\sqrt{4}*\sqrt{3102}=2\sqrt{3102}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3102}}{2*22}=\frac{0-2\sqrt{3102}}{44}
=-\frac{2\sqrt{3102}}{44}
=-\frac{\sqrt{3102}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3102}}{2*22}=\frac{0+2\sqrt{3102}}{44}
=\frac{2\sqrt{3102}}{44}
=\frac{\sqrt{3102}}{22}
$