You have to solve the given expression for q:
[tex]2q^2-8=3q[/tex]
This expression has a quadratic term, which means that it is a quadratic equation. To find the value or values of q, you have to use the quadratic equation, using "q" as the variable instead of "x"
[tex]q=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Where
a is the coefficient of the quadratic term
b is the coefficient of the q term
c is the constant
- First, zero the equation by passing 3q to the left side of the equal sign:
[tex]\begin{gathered} 2q^2-8=3q \\ 2q^2-8-3q=3q-3q \\ 2q^2-3q-8=0 \end{gathered}[/tex]
For this equation the coefficients are:
a= 2
b= -3
c= -8
Replace them in the formula and solve:
[tex]\begin{gathered} q=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ q=\frac{-(-3)\pm\sqrt{(-3)^2-4*2*(-8)}}{2*2} \\ q=\frac{3\pm\sqrt{9+64}}{4} \\ q=\frac{3\pm\sqrt{73}}{4} \end{gathered}[/tex]
Next, to determine each possible value of q, you have to solve the sum and difference separately:
Sum
[tex]\begin{gathered} q=\frac{3+\sqrt{73}}{4} \\ q=2.886\cong2.9 \end{gathered}[/tex]
Difference
[tex]\begin{gathered} q=\frac{3-\sqrt{73}}{4} \\ q=-1.386\cong-1.4 \end{gathered}[/tex]
The possible solutions for the given equation are q=2.9 and q=-1.4
