Solve 32x^2-80x+48=0

$$32 { x }^{ 2 } -80x+48=0$$

Solve for x

$$x=1$$

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

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$32x^{2}-80x+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 32 for a, -80 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 32\times 48}}{2\times 32}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 32\times 48}}{2\times 32}$$

Multiply -4 times 32.

$$x=\frac{-\left(-80\right)±\sqrt{6400-128\times 48}}{2\times 32}$$

Multiply -128 times 48.

$$x=\frac{-\left(-80\right)±\sqrt{6400-6144}}{2\times 32}$$

Add 6400 to -6144.

$$x=\frac{-\left(-80\right)±\sqrt{256}}{2\times 32}$$

Take the square root of 256.

$$x=\frac{-\left(-80\right)±16}{2\times 32}$$

The opposite of -80 is 80.

$$x=\frac{80±16}{2\times 32}$$

Multiply 2 times 32.

$$x=\frac{80±16}{64}$$

Now solve the equation x=\frac{80±16}{64} when ± is plus. Add 80 to 16.

$$x=\frac{96}{64}$$

Reduce the fraction \frac{96}{64} to lowest terms by extracting and canceling out 32.

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

Now solve the equation x=\frac{80±16}{64} when ± is minus. Subtract 16 from 80.

$$x=\frac{64}{64}$$

Divide 64 by 64.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$32x^{2}-80x+48=0$$

Subtract 48 from both sides of the equation.

$$32x^{2}-80x+48-48=-48$$

Subtracting 48 from itself leaves 0.

$$32x^{2}-80x=-48$$

Divide both sides by 32.

$$\frac{32x^{2}-80x}{32}=-\frac{48}{32}$$

Dividing by 32 undoes the multiplication by 32.

$$x^{2}+\left(-\frac{80}{32}\right)x=-\frac{48}{32}$$

Reduce the fraction \frac{-80}{32} to lowest terms by extracting and canceling out 16.

$$x^{2}-\frac{5}{2}x=-\frac{48}{32}$$

Reduce the fraction \frac{-48}{32} to lowest terms by extracting and canceling out 16.

$$x^{2}-\frac{5}{2}x=-\frac{3}{2}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{3}{2}+\frac{25}{16}$$

Add -\frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{1}{4} x-\frac{5}{4}=-\frac{1}{4}$$

Add \frac{5}{4} to both sides of the equation.

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