Solve left(4xright)^2+8x-3=0

$${ \left(4x \right) }^{ 2 } +8x-3 = 0$$

Solve for x

$$x=-\frac{3}{4}=-0.75$$

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

Steps Using Factoring By Grouping

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+8x-3=0$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+8x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=8 ab=16\left(-3\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

Rewrite 16x^{2}+8x-3 as \left(16x^{2}-4x\right)+\left(12x-3\right).

$$\left(16x^{2}-4x\right)+\left(12x-3\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(4x-1\right)+3\left(4x-1\right)$$

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

$$\left(4x-1\right)\left(4x+3\right)$$

To find equation solutions, solve 4x-1=0 and 4x+3=0.

$$x=\frac{1}{4} x=-\frac{3}{4}$$

Steps Using the Quadratic Formula

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+8x-3=0$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+8x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 8 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16\left(-3\right)}}{2\times 16}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 16\left(-3\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-8±\sqrt{64-64\left(-3\right)}}{2\times 16}$$

Multiply -64 times -3.

$$x=\frac{-8±\sqrt{64+192}}{2\times 16}$$

Add 64 to 192.

$$x=\frac{-8±\sqrt{256}}{2\times 16}$$

Take the square root of 256.

$$x=\frac{-8±16}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-8±16}{32}$$

Now solve the equation x=\frac{-8±16}{32} when ± is plus. Add -8 to 16.

$$x=\frac{8}{32}$$

Reduce the fraction \frac{8}{32} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-8±16}{32} when ± is minus. Subtract 16 from -8.

$$x=-\frac{24}{32}$$

Reduce the fraction \frac{-24}{32} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$x=\frac{1}{4} x=-\frac{3}{4}$$

Steps for Completing the Square

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+8x-3=0$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+8x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$16x^{2}+8x=3$$

Divide both sides by 16.

$$\frac{16x^{2}+8x}{16}=\frac{3}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{8}{16}x=\frac{3}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{1}{2}x=\frac{3}{16}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{3}{16}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{3+1}{16}$$

Add \frac{3}{16} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{4}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{1}{4} x=-\frac{3}{4}$$