Write the following equation in standard form. Then solve.6 q squared minus 5 qequals5 q squared minus 7 q plus 48Question content area bottomPart 1The equation in standard form isenter your response here

Answer 1

To rewrite the equation $6q^2-5q=5q^2-7q+48$ in standard form, we first move all terms to one side of the equation.

Start with the original equation:

$$6q^2-5q=5q^2-7q+48$$

Subtract $5q^2$, add $7q$, and subtract $48$ from both sides:

$$6q^2-5q-5q^2+7q-48=0$$

Now, combine like terms:

$$(6q^2-5q^2)+(-5q+7q)-48=0$$ $$q^2+2q-48=0$$

This is the equation in standard form.

Now, we can solve the quadratic equation $q^2+2q-48=0$ using the quadratic formula:

$$q=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Here, $a=1$, $b=2$, and $c=-48$.

First, calculate the discriminant:

$$b^2-4ac=2^2-4(1)(-48)=4+192=196$$

Now apply the quadratic formula:

$$q=\frac{-2\pm \sqrt{196}}{2(1)}$$ $$q=\frac{-2\pm 14}{2}$$

Now, we find the two possible values for $q$:

$$q=\frac{-2+14}{2}=\frac{12}{2}=6$$

$$q=\frac{-2-14}{2}=\frac{-16}{2}=-8$$

Thus, the solutions of the equation $q^2+2q-48=0$ are:

Answer:

The equation in standard form is $q^2+2q-48=0$.

The solutions are $q=6$ and $q=-8$.