Solve 27x-x^2+54-2x=210

$$27 x - x ^ { 2 } + 54 - 2 x = 210$$

Solve for x

$$x=12$$

$$x=13$$

Steps Using Factoring By Grouping

Combine 27x and -2x to get 25x.

$$25x-x^{2}+54=210$$

Subtract 210 from both sides.

$$25x-x^{2}+54-210=0$$

Subtract 210 from 54 to get -156.

$$25x-x^{2}-156=0$$

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

$$-x^{2}+25x-156=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-156. To find a and b, set up a system to be solved.

$$a+b=25 ab=-\left(-156\right)=156$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 156.

$$1,156 2,78 3,52 4,39 6,26 12,13$$

Calculate the sum for each pair.

$$1+156=157 2+78=80 3+52=55 4+39=43 6+26=32 12+13=25$$

The solution is the pair that gives sum 25.

$$a=13 b=12$$

Rewrite -x^{2}+25x-156 as \left(-x^{2}+13x\right)+\left(12x-156\right).

$$\left(-x^{2}+13x\right)+\left(12x-156\right)$$

Factor out -x in the first and 12 in the second group.

$$-x\left(x-13\right)+12\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(-x+12\right)$$

To find equation solutions, solve x-13=0 and -x+12=0.

$$x=13 x=12$$

Steps Using the Quadratic Formula

Combine 27x and -2x to get 25x.

$$25x-x^{2}+54=210$$

Subtract 210 from both sides.

$$25x-x^{2}+54-210=0$$

Subtract 210 from 54 to get -156.

$$25x-x^{2}-156=0$$

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.

$$-x^{2}+25x-156=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 25 for b, and -156 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-1\right)\left(-156\right)}}{2\left(-1\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-1\right)\left(-156\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-25±\sqrt{625+4\left(-156\right)}}{2\left(-1\right)}$$

Multiply 4 times -156.

$$x=\frac{-25±\sqrt{625-624}}{2\left(-1\right)}$$

Add 625 to -624.

$$x=\frac{-25±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-25±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-25±1}{-2}$$

Now solve the equation x=\frac{-25±1}{-2} when ± is plus. Add -25 to 1.

$$x=-\frac{24}{-2}$$

Divide -24 by -2.

$$x=12$$

Now solve the equation x=\frac{-25±1}{-2} when ± is minus. Subtract 1 from -25.

$$x=-\frac{26}{-2}$$

Divide -26 by -2.

$$x=13$$

The equation is now solved.

$$x=12 x=13$$

Steps for Completing the Square

Combine 27x and -2x to get 25x.

$$25x-x^{2}+54=210$$

Subtract 54 from both sides.

$$25x-x^{2}=210-54$$

Subtract 54 from 210 to get 156.

$$25x-x^{2}=156$$

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.

$$-x^{2}+25x=156$$

Divide both sides by -1.

$$\frac{-x^{2}+25x}{-1}=\frac{156}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{25}{-1}x=\frac{156}{-1}$$

Divide 25 by -1.

$$x^{2}-25x=\frac{156}{-1}$$

Divide 156 by -1.

$$x^{2}-25x=-156$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=-156+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-25x+\frac{625}{4}=-156+\frac{625}{4}$$

Add -156 to \frac{625}{4}.

$$x^{2}-25x+\frac{625}{4}=\frac{1}{4}$$

Factor x^{2}-25x+\frac{625}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{25}{2}=\frac{1}{2} x-\frac{25}{2}=-\frac{1}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$x=13 x=12$$