21x^2+32x-5=1

Solution for 21x^2+32x-5=1 equation:

21x^2+32x-5=1

We move all terms to the left:

21x^2+32x-5-(1)=0

We add all the numbers together, and all the variables

21x^2+32x-6=0

a = 21; b = 32; c = -6;
Δ = b2-4ac
Δ = 322-4·21·(-6)
Δ = 1528
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:

$\sqrt{\Delta}=\sqrt{1528}=\sqrt{4*382}=\sqrt{4}*\sqrt{382}=2\sqrt{382}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{382}}{2*21}=\frac{-32-2\sqrt{382}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{382}}{2*21}=\frac{-32+2\sqrt{382}}{42}
$