(-13/65)q+100=0

Solution for (-13/65)q+100=0 equation:

(-13/65)q+100=0


Domain of the equation: 65)q!=0

q!=0/1

q!=0

q∈R


We multiply parentheses

-13q^2+100=0

a = -13; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-13)·100
Δ = 5200
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:

$\sqrt{\Delta}=\sqrt{5200}=\sqrt{400*13}=\sqrt{400}*\sqrt{13}=20\sqrt{13}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{13}}{2*-13}=\frac{0-20\sqrt{13}}{-26}
=-\frac{20\sqrt{13}}{-26}
=-\frac{10\sqrt{13}}{-13}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{13}}{2*-13}=\frac{0+20\sqrt{13}}{-26}
=\frac{20\sqrt{13}}{-26}
=\frac{10\sqrt{13}}{-13}
$