Find a,b,c 7x^2-3x+8=9x^2+15

$7 x^{2} - 3 x + 8 = 9 x^{2} + 15$

Step 1

Move all terms to the left side of the equation and simplify.

Move all the expressions to the left side of the equation.

Subtract $9 x^{2}$ from both sides of the equation.

$7 x^{2} - 3 x + 8 - 9 x^{2} = 15$

Subtract $15$ from both sides of the equation.

$7 x^{2} - 3 x + 8 - 9 x^{2} - 15 = 0$

Simplify $7 x^{2} - 3 x + 8 - 9 x^{2} - 15$ .

Subtract $9 x^{2}$ from $7 x^{2}$ .

$- 2 x^{2} - 3 x + 8 - 15 = 0$

Subtract $15$ from $8$ .

$- 2 x^{2} - 3 x - 7 = 0$

Step 2

Once the quadratic is in standard form, the values of $a$ , $b$ , and $c$ can be found.

$a x^{2} + b x + c$

Step 3

Use the standard form of the equation to find $a$ , $b$ , and $c$ for this quadratic.

$a = - 2$ , $b = - 3$ , $c = - 7$