Question

How do i solve x^2+7x=3x

Answer 1

To solve, you must make one side of the equation equal to zero. We can do this by subtracting 3x on both sides of the equation.

x^2 + 7x = 3x
       - 3x    -3x

x^2 + 4x = 0.

Then we substitute the equation into the quadratic equation.

$\frac{-b+- \sqrt{b^2-4ac} }{2a}$

Where a = 1 b = 4 and c = 0

$\frac{-(4)+- \sqrt{(4)^2-4(1)(0)} }{2(1)}$

We first solve whats inside the square root.

$\frac{-(4)+- \sqrt{16} }{2(1)}$

$\sqrt{16} = 4$

Now we solve for -4 and 4.

$-4 - 4 = -8$
$-4+4=0$

Then divide both answer by 2

$\frac{-8}{2} = -4$
The division of any by zero is zero

So the answer are x = -4 and x = 0

Answer 2

X^2 + 7x - 3x = 0
Simplify x^2 + 7x - 3x to x^2 + 4x
X^2 + 4x = 0
Factor out the common term x
X ( x + 4 ) = 0
Solve for x
X = 0,-4