4r^2-12r+7=0

Solution for 4r^2-12r+7=0 equation:

4r^2-12r+7=0

a = 4; b = -12; c = +7;
Δ = b2-4ac
Δ = -122-4·4·7
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{2}}{2*4}=\frac{12-4\sqrt{2}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{2}}{2*4}=\frac{12+4\sqrt{2}}{8}
$