7x^2-38x+27=0

Solution for 7x^2-38x+27=0 equation:

7x^2-38x+27=0

a = 7; b = -38; c = +27;
Δ = b2-4ac
Δ = -382-4·7·27
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-4\sqrt{43}}{2*7}=\frac{38-4\sqrt{43}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+4\sqrt{43}}{2*7}=\frac{38+4\sqrt{43}}{14}
$