16x^2+40x+6=27

Solution for 16x^2+40x+6=27 equation:

16x^2+40x+6=27

We move all terms to the left:

16x^2+40x+6-(27)=0

We add all the numbers together, and all the variables

16x^2+40x-21=0

a = 16; b = 40; c = -21;
Δ = b2-4ac
Δ = 402-4·16·(-21)
Δ = 2944
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:

$\sqrt{\Delta}=\sqrt{2944}=\sqrt{64*46}=\sqrt{64}*\sqrt{46}=8\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{46}}{2*16}=\frac{-40-8\sqrt{46}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{46}}{2*16}=\frac{-40+8\sqrt{46}}{32}
$