4r^2-94r+345=0

Solution for 4r^2-94r+345=0 equation:

4r^2-94r+345=0

a = 4; b = -94; c = +345;
Δ = b2-4ac
Δ = -942-4·4·345
Δ = 3316
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:

$\sqrt{\Delta}=\sqrt{3316}=\sqrt{4*829}=\sqrt{4}*\sqrt{829}=2\sqrt{829}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-94)-2\sqrt{829}}{2*4}=\frac{94-2\sqrt{829}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-94)+2\sqrt{829}}{2*4}=\frac{94+2\sqrt{829}}{8}
$