3x^2+14x-5=1

Solution for 3x^2+14x-5=1 equation:

3x^2+14x-5=1

We move all terms to the left:

3x^2+14x-5-(1)=0

We add all the numbers together, and all the variables

3x^2+14x-6=0

a = 3; b = 14; c = -6;
Δ = b2-4ac
Δ = 142-4·3·(-6)
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{67}}{2*3}=\frac{-14-2\sqrt{67}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{67}}{2*3}=\frac{-14+2\sqrt{67}}{6}
$