0.12x^2+0.9x+2.9x=10

Solution for 0.12x^2+0.9x+2.9x=10 equation:

0.12x^2+0.9x+2.9x=10

We move all terms to the left:

0.12x^2+0.9x+2.9x-(10)=0

We add all the numbers together, and all the variables

0.12x^2+3.8x-10=0

a = 0.12; b = 3.8; c = -10;
Δ = b2-4ac
Δ = 3.82-4·0.12·(-10)
Δ = 19.24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3.8)-\sqrt{19.24}}{2*0.12}=\frac{-3.8-\sqrt{19.24}}{0.24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.8)+\sqrt{19.24}}{2*0.12}=\frac{-3.8+\sqrt{19.24}}{0.24}
$