16x+3=29-3x;2

Solution for 16x+3=29-3x;2 equation:

16x+3=29-3x^2

We move all terms to the left:

16x+3-(29-3x^2)=0

We get rid of parentheses

3x^2+16x-29+3=0

We add all the numbers together, and all the variables

3x^2+16x-26=0

a = 3; b = 16; c = -26;
Δ = b2-4ac
Δ = 162-4·3·(-26)
Δ = 568
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:

$\sqrt{\Delta}=\sqrt{568}=\sqrt{4*142}=\sqrt{4}*\sqrt{142}=2\sqrt{142}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{142}}{2*3}=\frac{-16-2\sqrt{142}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{142}}{2*3}=\frac{-16+2\sqrt{142}}{6}
$