Solve 6x^2+60=39x

$$6 x ^ { 2 } + 60 = 39 x$$

Solve for x

$$x=4$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Subtract 39x from both sides.

$$6x^{2}+60-39x=0$$

Divide both sides by 3.

$$2x^{2}+20-13x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}-13x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\times 20=40$$

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 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite 2x^{2}-13x+20 as \left(2x^{2}-8x\right)+\left(-5x+20\right).

$$\left(2x^{2}-8x\right)+\left(-5x+20\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-4\right)-5\left(x-4\right)$$

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

$$\left(x-4\right)\left(2x-5\right)$$

To find equation solutions, solve x-4=0 and 2x-5=0.

$$x=4 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

Subtract 39x from both sides.

$$6x^{2}+60-39x=0$$

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.

$$6x^{2}-39x+60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -39 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 6\times 60}}{2\times 6}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\times 6\times 60}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-39\right)±\sqrt{1521-24\times 60}}{2\times 6}$$

Multiply -24 times 60.

$$x=\frac{-\left(-39\right)±\sqrt{1521-1440}}{2\times 6}$$

Add 1521 to -1440.

$$x=\frac{-\left(-39\right)±\sqrt{81}}{2\times 6}$$

Take the square root of 81.

$$x=\frac{-\left(-39\right)±9}{2\times 6}$$

The opposite of -39 is 39.

$$x=\frac{39±9}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{39±9}{12}$$

Now solve the equation x=\frac{39±9}{12} when ± is plus. Add 39 to 9.

$$x=\frac{48}{12}$$

Divide 48 by 12.

$$x=4$$

Now solve the equation x=\frac{39±9}{12} when ± is minus. Subtract 9 from 39.

$$x=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{2}$$

The equation is now solved.

$$x=4 x=\frac{5}{2}$$

Steps for Completing the Square

Subtract 39x from both sides.

$$6x^{2}+60-39x=0$$

Subtract 60 from both sides. Anything subtracted from zero gives its negation.

$$6x^{2}-39x=-60$$

Divide both sides by 6.

$$\frac{6x^{2}-39x}{6}=-\frac{60}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{39}{6}\right)x=-\frac{60}{6}$$

Reduce the fraction \frac{-39}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{13}{2}x=-\frac{60}{6}$$

Divide -60 by 6.

$$x^{2}-\frac{13}{2}x=-10$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-10+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-10+\frac{169}{16}$$

Add -10 to \frac{169}{16}.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{3}{4} x-\frac{13}{4}=-\frac{3}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=4 x=\frac{5}{2}$$