10x2+68x=14

Solution for 10x2+68x=14 equation:

10x^2+68x=14

We move all terms to the left:

10x^2+68x-(14)=0

a = 10; b = 68; c = -14;
Δ = b2-4ac
Δ = 682-4·10·(-14)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-72}{2*10}=\frac{-140}{20}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+72}{2*10}=\frac{4}{20}
=1/5
$