6x^2+16x-18=1

Solution for 6x^2+16x-18=1 equation:

6x^2+16x-18=1

We move all terms to the left:

6x^2+16x-18-(1)=0

We add all the numbers together, and all the variables

6x^2+16x-19=0

a = 6; b = 16; c = -19;
Δ = b2-4ac
Δ = 162-4·6·(-19)
Δ = 712
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:

$\sqrt{\Delta}=\sqrt{712}=\sqrt{4*178}=\sqrt{4}*\sqrt{178}=2\sqrt{178}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{178}}{2*6}=\frac{-16-2\sqrt{178}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{178}}{2*6}=\frac{-16+2\sqrt{178}}{12}
$