Solve 4x^2-53x+60=0

$$4 { x }^{ 2 } -53x+60 = 0$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$x=12$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+60. To find a and b, set up a system to be solved.

$$a+b=-53 ab=4\times 60=240$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 240.

$$-1,-240 -2,-120 -3,-80 -4,-60 -5,-48 -6,-40 -8,-30 -10,-24 -12,-20 -15,-16$$

Calculate the sum for each pair.

$$-1-240=-241 -2-120=-122 -3-80=-83 -4-60=-64 -5-48=-53 -6-40=-46 -8-30=-38 -10-24=-34 -12-20=-32 -15-16=-31$$

The solution is the pair that gives sum -53.

$$a=-48 b=-5$$

Rewrite 4x^{2}-53x+60 as \left(4x^{2}-48x\right)+\left(-5x+60\right).

$$\left(4x^{2}-48x\right)+\left(-5x+60\right)$$

Factor out 4x in the first and -5 in the second group.

$$4x\left(x-12\right)-5\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(4x-5\right)$$

To find equation solutions, solve x-12=0 and 4x-5=0.

$$x=12 x=\frac{5}{4}$$

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.

$$4x^{2}-53x+60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -53 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-53\right)±\sqrt{\left(-53\right)^{2}-4\times 4\times 60}}{2\times 4}$$

Square -53.

$$x=\frac{-\left(-53\right)±\sqrt{2809-4\times 4\times 60}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-53\right)±\sqrt{2809-16\times 60}}{2\times 4}$$

Multiply -16 times 60.

$$x=\frac{-\left(-53\right)±\sqrt{2809-960}}{2\times 4}$$

Add 2809 to -960.

$$x=\frac{-\left(-53\right)±\sqrt{1849}}{2\times 4}$$

Take the square root of 1849.

$$x=\frac{-\left(-53\right)±43}{2\times 4}$$

The opposite of -53 is 53.

$$x=\frac{53±43}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{53±43}{8}$$

Now solve the equation x=\frac{53±43}{8} when ± is plus. Add 53 to 43.

$$x=\frac{96}{8}$$

Divide 96 by 8.

$$x=12$$

Now solve the equation x=\frac{53±43}{8} when ± is minus. Subtract 43 from 53.

$$x=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=12 x=\frac{5}{4}$$

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.

$$4x^{2}-53x+60=0$$

Subtract 60 from both sides of the equation.

$$4x^{2}-53x+60-60=-60$$

Subtracting 60 from itself leaves 0.

$$4x^{2}-53x=-60$$

Divide both sides by 4.

$$\frac{4x^{2}-53x}{4}=-\frac{60}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{53}{4}x=-\frac{60}{4}$$

Divide -60 by 4.

$$x^{2}-\frac{53}{4}x=-15$$

Divide -\frac{53}{4}, the coefficient of the x term, by 2 to get -\frac{53}{8}. Then add the square of -\frac{53}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{53}{4}x+\left(-\frac{53}{8}\right)^{2}=-15+\left(-\frac{53}{8}\right)^{2}$$

Square -\frac{53}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{53}{4}x+\frac{2809}{64}=-15+\frac{2809}{64}$$

Add -15 to \frac{2809}{64}.

$$x^{2}-\frac{53}{4}x+\frac{2809}{64}=\frac{1849}{64}$$

Factor x^{2}-\frac{53}{4}x+\frac{2809}{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}.

$$\left(x-\frac{53}{8}\right)^{2}=\frac{1849}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{53}{8}\right)^{2}}=\sqrt{\frac{1849}{64}}$$

Simplify.

$$x-\frac{53}{8}=\frac{43}{8} x-\frac{53}{8}=-\frac{43}{8}$$

Add \frac{53}{8} to both sides of the equation.

$$x=12 x=\frac{5}{4}$$