30x^2+40x=1

Solution for 30x^2+40x=1 equation:

30x^2+40x=1

We move all terms to the left:

30x^2+40x-(1)=0

a = 30; b = 40; c = -1;
Δ = b2-4ac
Δ = 402-4·30·(-1)
Δ = 1720
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:

$\sqrt{\Delta}=\sqrt{1720}=\sqrt{4*430}=\sqrt{4}*\sqrt{430}=2\sqrt{430}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{430}}{2*30}=\frac{-40-2\sqrt{430}}{60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{430}}{2*30}=\frac{-40+2\sqrt{430}}{60}
$