5x2+27x=10

Solution for 5x2+27x=10 equation:

5x^2+27x=10

We move all terms to the left:

5x^2+27x-(10)=0

a = 5; b = 27; c = -10;
Δ = b2-4ac
Δ = 272-4·5·(-10)
Δ = 929
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{-(27)-\sqrt{929}}{2*5}=\frac{-27-\sqrt{929}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{929}}{2*5}=\frac{-27+\sqrt{929}}{10}
$