Solve 4x^2+3x+1=3-4x

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

Solve for x

$$x=-2$$

$$x=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

Subtract 3 from both sides.

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

Subtract 3 from 1 to get -2.

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

Add 4x to both sides.

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

Combine 3x and 4x to get 7x.

$$4x^{2}+7x-2=0$$

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

$$a+b=7 ab=4\left(-2\right)=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=-1 b=8$$

Rewrite 4x^{2}+7x-2 as \left(4x^{2}-x\right)+\left(8x-2\right).

$$\left(4x^{2}-x\right)+\left(8x-2\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(4x-1\right)+2\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(x+2\right)$$

To find equation solutions, solve 4x-1=0 and x+2=0.

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

Steps Using the Quadratic Formula

Subtract 3 from both sides.

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

Subtract 3 from 1 to get -2.

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

Add 4x to both sides.

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

Combine 3x and 4x to get 7x.

$$4x^{2}+7x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 7 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 4\left(-2\right)}}{2\times 4}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 4\left(-2\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-7±\sqrt{49-16\left(-2\right)}}{2\times 4}$$

Multiply -16 times -2.

$$x=\frac{-7±\sqrt{49+32}}{2\times 4}$$

Add 49 to 32.

$$x=\frac{-7±\sqrt{81}}{2\times 4}$$

Take the square root of 81.

$$x=\frac{-7±9}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-7±9}{8}$$

Now solve the equation x=\frac{-7±9}{8} when ± is plus. Add -7 to 9.

$$x=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-7±9}{8} when ± is minus. Subtract 9 from -7.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Add 4x to both sides.

$$4x^{2}+3x+1+4x=3$$

Combine 3x and 4x to get 7x.

$$4x^{2}+7x+1=3$$

Subtract 1 from both sides.

$$4x^{2}+7x=3-1$$

Subtract 1 from 3 to get 2.

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

Divide both sides by 4.

$$\frac{4x^{2}+7x}{4}=\frac{2}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{7}{4}x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{7}{4}x=\frac{1}{2}$$

Divide \frac{7}{4}, the coefficient of the x term, by 2 to get \frac{7}{8}. Then add the square of \frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{4}x+\left(\frac{7}{8}\right)^{2}=\frac{1}{2}+\left(\frac{7}{8}\right)^{2}$$

Square \frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{1}{2}+\frac{49}{64}$$

Add \frac{1}{2} to \frac{49}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{81}{64}$$

Factor x^{2}+\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$x+\frac{7}{8}=\frac{9}{8} x+\frac{7}{8}=-\frac{9}{8}$$

Subtract \frac{7}{8} from both sides of the equation.

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