3x^2+4x+12+2x^2-17x+92=110

Solution for 3x^2+4x+12+2x^2-17x+92=110 equation:

3x^2+4x+12+2x^2-17x+92=110

We move all terms to the left:

3x^2+4x+12+2x^2-17x+92-(110)=0

We add all the numbers together, and all the variables

5x^2-13x-6=0

a = 5; b = -13; c = -6;
Δ = b2-4ac
Δ = -132-4·5·(-6)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*5}=\frac{-4}{10}
=-2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*5}=\frac{30}{10}
=3
$