3/2q+8=q=10

Solution for 3/2q+8=q=10 equation:

3/2q+8=q=10

We move all terms to the left:

3/2q+8-(q)=0


Domain of the equation: 2q!=0

q!=0/2

q!=0

q∈R


We add all the numbers together, and all the variables

-1q+3/2q+8=0

We multiply all the terms by the denominator


-1q*2q+8*2q+3=0

Wy multiply elements

-2q^2+16q+3=0

a = -2; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·(-2)·3
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{70}}{2*-2}=\frac{-16-2\sqrt{70}}{-4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{70}}{2*-2}=\frac{-16+2\sqrt{70}}{-4}
$