Solve for c 8x+4-(6x^2+x-3)=ax^2+bx+c

$8 x + 4 - (6 x^{2} + x - 3) = a x^{2} + b x + c$

Step 1

Rewrite the equation as $a x^{2} + b x + c = 8 x + 4 - (6 x^{2} + x - 3)$ .

$a x^{2} + b x + c = 8 x + 4 - (6 x^{2} + x - 3)$

Step 2

Simplify $8 x + 4 - (6 x^{2} + x - 3)$ .

Simplify each term.

Apply the distributive property.

$a x^{2} + b x + c = 8 x + 4 - (6 x^{2}) - x - - 3$

Simplify.

Multiply $6$ by $- 1$ .

$a x^{2} + b x + c = 8 x + 4 - 6 x^{2} - x - - 3$

Multiply $- 1$ by $- 3$ .

$a x^{2} + b x + c = 8 x + 4 - 6 x^{2} - x + 3$

Simplify by adding terms.

Subtract $x$ from $8 x$ .

$a x^{2} + b x + c = 7 x + 4 - 6 x^{2} + 3$

Add $4$ and $3$ .

$a x^{2} + b x + c = 7 x - 6 x^{2} + 7$

Step 3

Move all terms not containing $c$ to the right side of the equation.

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

$b x + c = 7 x - 6 x^{2} + 7 - a x^{2}$

Subtract $b x$ from both sides of the equation.

$c = 7 x - 6 x^{2} + 7 - a x^{2} - b x$