(3x)^2+8x-1=10

Solution for (3x)^2+8x-1=10 equation:

(3x)^2+8x-1=10

We move all terms to the left:

(3x)^2+8x-1-(10)=0

We add all the numbers together, and all the variables

3x^2+8x-11=0

a = 3; b = 8; c = -11;
Δ = b2-4ac
Δ = 82-4·3·(-11)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-14}{2*3}=\frac{-22}{6}
=-3+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*3}=\frac{6}{6}
=1
$