3q+15=(1/2)(q-5)

Solution for 3q+15=(1/2)(q-5) equation:

3q+15=(1/2)(q-5)

We move all terms to the left:

3q+15-((1/2)(q-5))=0


Domain of the equation: 2)(q-5))!=0

q∈R


We add all the numbers together, and all the variables

3q-((+1/2)(q-5))+15=0

We multiply parentheses ..

-((+q^2+1/2*-5))+3q+15=0

We multiply all the terms by the denominator


-((+q^2+1+3q*2*-5))+15*2*-5))=0


We calculate terms in parentheses: -((+q^2+1+3q*2*-5)), so:

(+q^2+1+3q*2*-5)

We get rid of parentheses

q^2+3q*2*+1-5

We add all the numbers together, and all the variables

q^2+3q*2*-4

Wy multiply elements

q^2+6q^2-4

We add all the numbers together, and all the variables

7q^2-4

Back to the equation:

-(7q^2-4)


We add all the numbers together, and all the variables

-(7q^2-4)=0

We get rid of parentheses

-7q^2+4=0

a = -7; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-7)·4
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{7}}{2*-7}=\frac{0-4\sqrt{7}}{-14}
=-\frac{4\sqrt{7}}{-14}
=-\frac{2\sqrt{7}}{-7}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{7}}{2*-7}=\frac{0+4\sqrt{7}}{-14}
=\frac{4\sqrt{7}}{-14}
=\frac{2\sqrt{7}}{-7}
$