Solve 4x^2+3x=7x+3

$$4 x ^ { 2 } + 3 x = 7 x + 3$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

$$4x^{2}+3x-7x=3$$

Combine 3x and -7x to get -4x.

$$4x^{2}-4x=3$$

Subtract 3 from both sides.

$$4x^{2}-4x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=4\left(-3\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite 4x^{2}-4x-3 as \left(4x^{2}-6x\right)+\left(2x-3\right).

$$\left(4x^{2}-6x\right)+\left(2x-3\right)$$

Factor out 2x in 4x^{2}-6x.

$$2x\left(2x-3\right)+2x-3$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(2x+1\right)$$

To find equation solutions, solve 2x-3=0 and 2x+1=0.

$$x=\frac{3}{2} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$4x^{2}+3x-7x=3$$

Combine 3x and -7x to get -4x.

$$4x^{2}-4x=3$$

Subtract 3 from both sides.

$$4x^{2}-4x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-3\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 4}$$

Add 16 to 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{-\left(-4\right)±8}{2\times 4}$$

The opposite of -4 is 4.

$$x=\frac{4±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±8}{8}$$

Now solve the equation x=\frac{4±8}{8} when ± is plus. Add 4 to 8.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{4±8}{8} when ± is minus. Subtract 8 from 4.

$$x=-\frac{4}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=\frac{3}{2} x=-\frac{1}{2}$$

Steps for Completing the Square

Subtract 7x from both sides.

$$4x^{2}+3x-7x=3$$

Combine 3x and -7x to get -4x.

$$4x^{2}-4x=3$$

Divide both sides by 4.

$$\frac{4x^{2}-4x}{4}=\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=\frac{3}{4}$$

Divide -4 by 4.

$$x^{2}-x=\frac{3}{4}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{3}{4}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{3+1}{4}$$

Add \frac{3}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=1$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-\frac{1}{2}=1 x-\frac{1}{2}=-1$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{3}{2} x=-\frac{1}{2}$$