7x^2+8x=1

Solution for 7x^2+8x=1 equation:

7x^2+8x=1

We move all terms to the left:

7x^2+8x-(1)=0

a = 7; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·7·(-1)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{23}}{2*7}=\frac{-8-2\sqrt{23}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{23}}{2*7}=\frac{-8+2\sqrt{23}}{14}
$