# Algebra Help

How would I solve this equation:

3(x-3) + 4x+7= 5x-3

And How would I solve these Quadratic Equation:

x^2 + 7x + 12 = 0

3x^2 - 10x + 8 = 0

8y^2 + 18y = 5 = 0

Thanks.

For the first one, distribute $$3(x-3)$$ and the group the x's on one side of the equation, and the rest of the numbers on the other side.

For the quadratics, try factoring. If you don't want to use that method, you can always use the quadratic formula.

Jameson

TD
Homework Helper
Nicholasw said:
3(x-3) + 4x+7= 5x-3
Simplify it by working out the paranthesis and put everything in x on 1 side, this should give an easy lineair equation.

Nicholasw said:
x^2 + 7x + 12 = 0

3x^2 - 10x + 8 = 0

8y^2 + 18y = 5 = 0
Have you seen the quadratic formula to solve these solutions? If an equation is given in the form $ax^2 + bx + c = 0$, then the solutions are given by:

$$x_{1,2} = \frac{{ - b \pm \sqrt {b^2 - 4ac} }} {{2a}}$$