2x^2+7x-5=1

Solution for 2x^2+7x-5=1 equation:

2x^2+7x-5=1

We move all terms to the left:

2x^2+7x-5-(1)=0

We add all the numbers together, and all the variables

2x^2+7x-6=0

a = 2; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·2·(-6)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{97}}{2*2}=\frac{-7-\sqrt{97}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{97}}{2*2}=\frac{-7+\sqrt{97}}{4}
$