Solve x^2+5x+136=1000

$${ x }^{ 2 } +5x+136=1000$$

Solve for x

$$x=-32$$

$$x=27$$

Steps Using Factoring

Subtract 1000 from both sides.

$$x^{2}+5x+136-1000=0$$

Subtract 1000 from 136 to get -864.

$$x^{2}+5x-864=0$$

To solve the equation, factor x^{2}+5x-864 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=-864$$

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 -864.

$$-1,864 -2,432 -3,288 -4,216 -6,144 -8,108 -9,96 -12,72 -16,54 -18,48 -24,36 -27,32$$

Calculate the sum for each pair.

$$-1+864=863 -2+432=430 -3+288=285 -4+216=212 -6+144=138 -8+108=100 -9+96=87 -12+72=60 -16+54=38 -18+48=30 -24+36=12 -27+32=5$$

The solution is the pair that gives sum 5.

$$a=-27 b=32$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-27\right)\left(x+32\right)$$

To find equation solutions, solve x-27=0 and x+32=0.

$$x=27 x=-32$$

Steps Using Factoring By Grouping

Subtract 1000 from both sides.

$$x^{2}+5x+136-1000=0$$

Subtract 1000 from 136 to get -864.

$$x^{2}+5x-864=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-864. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-864\right)=-864$$

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 -864.

$$-1,864 -2,432 -3,288 -4,216 -6,144 -8,108 -9,96 -12,72 -16,54 -18,48 -24,36 -27,32$$

Calculate the sum for each pair.

$$-1+864=863 -2+432=430 -3+288=285 -4+216=212 -6+144=138 -8+108=100 -9+96=87 -12+72=60 -16+54=38 -18+48=30 -24+36=12 -27+32=5$$

The solution is the pair that gives sum 5.

$$a=-27 b=32$$

Rewrite x^{2}+5x-864 as \left(x^{2}-27x\right)+\left(32x-864\right).

$$\left(x^{2}-27x\right)+\left(32x-864\right)$$

Factor out x in the first and 32 in the second group.

$$x\left(x-27\right)+32\left(x-27\right)$$

Factor out common term x-27 by using distributive property.

$$\left(x-27\right)\left(x+32\right)$$

To find equation solutions, solve x-27=0 and x+32=0.

$$x=27 x=-32$$

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.

$$x^{2}+5x+136=1000$$

Subtract 1000 from both sides of the equation.

$$x^{2}+5x+136-1000=1000-1000$$

Subtracting 1000 from itself leaves 0.

$$x^{2}+5x+136-1000=0$$

Subtract 1000 from 136.

$$x^{2}+5x-864=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -864 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-864\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-864\right)}}{2}$$

Multiply -4 times -864.

$$x=\frac{-5±\sqrt{25+3456}}{2}$$

Add 25 to 3456.

$$x=\frac{-5±\sqrt{3481}}{2}$$

Take the square root of 3481.

$$x=\frac{-5±59}{2}$$

Now solve the equation x=\frac{-5±59}{2} when ± is plus. Add -5 to 59.

$$x=\frac{54}{2}$$

Divide 54 by 2.

$$x=27$$

Now solve the equation x=\frac{-5±59}{2} when ± is minus. Subtract 59 from -5.

$$x=-\frac{64}{2}$$

Divide -64 by 2.

$$x=-32$$

The equation is now solved.

$$x=27 x=-32$$

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.

$$x^{2}+5x+136=1000$$

Subtract 136 from both sides of the equation.

$$x^{2}+5x+136-136=1000-136$$

Subtracting 136 from itself leaves 0.

$$x^{2}+5x=1000-136$$

Subtract 136 from 1000.

$$x^{2}+5x=864$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=864+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=864+\frac{25}{4}$$

Add 864 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{3481}{4}$$

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

$$\left(x+\frac{5}{2}\right)^{2}=\frac{3481}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{3481}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{59}{2} x+\frac{5}{2}=-\frac{59}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=27 x=-32$$