9x^2+6x-4=67

Solution for 9x^2+6x-4=67 equation:

9x^2+6x-4=67

We move all terms to the left:

9x^2+6x-4-(67)=0

We add all the numbers together, and all the variables

9x^2+6x-71=0

a = 9; b = 6; c = -71;
Δ = b2-4ac
Δ = 62-4·9·(-71)
Δ = 2592
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:

$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-36\sqrt{2}}{2*9}=\frac{-6-36\sqrt{2}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+36\sqrt{2}}{2*9}=\frac{-6+36\sqrt{2}}{18}
$