10x^2+51x+27=0

Solution for 10x^2+51x+27=0 equation:

10x^2+51x+27=0

a = 10; b = 51; c = +27;
Δ = b2-4ac
Δ = 512-4·10·27
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-39}{2*10}=\frac{-90}{20}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+39}{2*10}=\frac{-12}{20}
=-3/5
$