112q^2=17q+5

Solution for 112q^2=17q+5 equation:

112q^2=17q+5

We move all terms to the left:

112q^2-(17q+5)=0

We get rid of parentheses

112q^2-17q-5=0

a = 112; b = -17; c = -5;
Δ = b2-4ac
Δ = -172-4·112·(-5)
Δ = 2529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2529}=\sqrt{9*281}=\sqrt{9}*\sqrt{281}=3\sqrt{281}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-3\sqrt{281}}{2*112}=\frac{17-3\sqrt{281}}{224}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+3\sqrt{281}}{2*112}=\frac{17+3\sqrt{281}}{224}
$