10q+3q+5=2(q-3)q=

Solution for 10q+3q+5=2(q-3)q= equation:

10q+3q+5=2(q-3)q=

We move all terms to the left:

10q+3q+5-(2(q-3)q)=0

We add all the numbers together, and all the variables

13q-(2(q-3)q)+5=0


We calculate terms in parentheses: -(2(q-3)q), so:

2(q-3)q

We multiply parentheses

2q^2-6q

Back to the equation:

-(2q^2-6q)


We get rid of parentheses

-2q^2+13q+6q+5=0

We add all the numbers together, and all the variables

-2q^2+19q+5=0

a = -2; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·(-2)·5
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{401}}{2*-2}=\frac{-19-\sqrt{401}}{-4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{401}}{2*-2}=\frac{-19+\sqrt{401}}{-4}
$