Solve 0=(-(2m+10))^2-4(1+m^2)(18)

$$0 = ( - ( 2 m + 10 ) ) ^ { 2 } - 4 ( 1 + m ^ { 2 } ) ( 18 )$$

Solve for m

$$m=-\frac{7}{17}\approx -0.411764706$$

$$m=1$$

Steps Using Factoring By Grouping

To find the opposite of 2m+10, find the opposite of each term.

$$0=\left(-2m-10\right)^{2}-4\left(1+m^{2}\right)\times 18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2m-10\right)^{2}.

$$0=4m^{2}+40m+100-4\left(1+m^{2}\right)\times 18$$

Multiply 4 and 18 to get 72.

$$0=4m^{2}+40m+100-72\left(1+m^{2}\right)$$

Use the distributive property to multiply -72 by 1+m^{2}.

$$0=4m^{2}+40m+100-72-72m^{2}$$

Subtract 72 from 100 to get 28.

$$0=4m^{2}+40m+28-72m^{2}$$

Combine 4m^{2} and -72m^{2} to get -68m^{2}.

$$0=-68m^{2}+40m+28$$

Swap sides so that all variable terms are on the left hand side.

$$-68m^{2}+40m+28=0$$

Divide both sides by 4.

$$-17m^{2}+10m+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -17m^{2}+am+bm+7. To find a and b, set up a system to be solved.

$$a+b=10 ab=-17\times 7=-119$$

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 -119.

$$-1,119 -7,17$$

Calculate the sum for each pair.

$$-1+119=118 -7+17=10$$

The solution is the pair that gives sum 10.

$$a=17 b=-7$$

Rewrite -17m^{2}+10m+7 as \left(-17m^{2}+17m\right)+\left(-7m+7\right).

$$\left(-17m^{2}+17m\right)+\left(-7m+7\right)$$

Factor out 17m in the first and 7 in the second group.

$$17m\left(-m+1\right)+7\left(-m+1\right)$$

Factor out common term -m+1 by using distributive property.

$$\left(-m+1\right)\left(17m+7\right)$$

To find equation solutions, solve -m+1=0 and 17m+7=0.

$$m=1 m=-\frac{7}{17}$$

Steps Using the Quadratic Formula

To find the opposite of 2m+10, find the opposite of each term.

$$0=\left(-2m-10\right)^{2}-4\left(1+m^{2}\right)\times 18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2m-10\right)^{2}.

$$0=4m^{2}+40m+100-4\left(1+m^{2}\right)\times 18$$

Multiply 4 and 18 to get 72.

$$0=4m^{2}+40m+100-72\left(1+m^{2}\right)$$

Use the distributive property to multiply -72 by 1+m^{2}.

$$0=4m^{2}+40m+100-72-72m^{2}$$

Subtract 72 from 100 to get 28.

$$0=4m^{2}+40m+28-72m^{2}$$

Combine 4m^{2} and -72m^{2} to get -68m^{2}.

$$0=-68m^{2}+40m+28$$

Swap sides so that all variable terms are on the left hand side.

$$-68m^{2}+40m+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -68 for a, 40 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-40±\sqrt{40^{2}-4\left(-68\right)\times 28}}{2\left(-68\right)}$$

Square 40.

$$m=\frac{-40±\sqrt{1600-4\left(-68\right)\times 28}}{2\left(-68\right)}$$

Multiply -4 times -68.

$$m=\frac{-40±\sqrt{1600+272\times 28}}{2\left(-68\right)}$$

Multiply 272 times 28.

$$m=\frac{-40±\sqrt{1600+7616}}{2\left(-68\right)}$$

Add 1600 to 7616.

$$m=\frac{-40±\sqrt{9216}}{2\left(-68\right)}$$

Take the square root of 9216.

$$m=\frac{-40±96}{2\left(-68\right)}$$

Multiply 2 times -68.

$$m=\frac{-40±96}{-136}$$

Now solve the equation m=\frac{-40±96}{-136} when ± is plus. Add -40 to 96.

$$m=\frac{56}{-136}$$

Reduce the fraction \frac{56}{-136} to lowest terms by extracting and canceling out 8.

$$m=-\frac{7}{17}$$

Now solve the equation m=\frac{-40±96}{-136} when ± is minus. Subtract 96 from -40.

$$m=-\frac{136}{-136}$$

Divide -136 by -136.

$$m=1$$

The equation is now solved.

$$m=-\frac{7}{17} m=1$$

Steps for Completing the Square

To find the opposite of 2m+10, find the opposite of each term.

$$0=\left(-2m-10\right)^{2}-4\left(1+m^{2}\right)\times 18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2m-10\right)^{2}.

$$0=4m^{2}+40m+100-4\left(1+m^{2}\right)\times 18$$

Multiply 4 and 18 to get 72.

$$0=4m^{2}+40m+100-72\left(1+m^{2}\right)$$

Use the distributive property to multiply -72 by 1+m^{2}.

$$0=4m^{2}+40m+100-72-72m^{2}$$

Subtract 72 from 100 to get 28.

$$0=4m^{2}+40m+28-72m^{2}$$

Combine 4m^{2} and -72m^{2} to get -68m^{2}.

$$0=-68m^{2}+40m+28$$

Swap sides so that all variable terms are on the left hand side.

$$-68m^{2}+40m+28=0$$

Subtract 28 from both sides. Anything subtracted from zero gives its negation.

$$-68m^{2}+40m=-28$$

Divide both sides by -68.

$$\frac{-68m^{2}+40m}{-68}=-\frac{28}{-68}$$

Dividing by -68 undoes the multiplication by -68.

$$m^{2}+\frac{40}{-68}m=-\frac{28}{-68}$$

Reduce the fraction \frac{40}{-68} to lowest terms by extracting and canceling out 4.

$$m^{2}-\frac{10}{17}m=-\frac{28}{-68}$$

Reduce the fraction \frac{-28}{-68} to lowest terms by extracting and canceling out 4.

$$m^{2}-\frac{10}{17}m=\frac{7}{17}$$

Divide -\frac{10}{17}, the coefficient of the x term, by 2 to get -\frac{5}{17}. Then add the square of -\frac{5}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-\frac{10}{17}m+\left(-\frac{5}{17}\right)^{2}=\frac{7}{17}+\left(-\frac{5}{17}\right)^{2}$$

Square -\frac{5}{17} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-\frac{10}{17}m+\frac{25}{289}=\frac{7}{17}+\frac{25}{289}$$

Add \frac{7}{17} to \frac{25}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}-\frac{10}{17}m+\frac{25}{289}=\frac{144}{289}$$

Factor m^{2}-\frac{10}{17}m+\frac{25}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-\frac{5}{17}\right)^{2}=\frac{144}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{5}{17}\right)^{2}}=\sqrt{\frac{144}{289}}$$

Simplify.

$$m-\frac{5}{17}=\frac{12}{17} m-\frac{5}{17}=-\frac{12}{17}$$

Add \frac{5}{17} to both sides of the equation.

$$m=1 m=-\frac{7}{17}$$