Solve (-4m_{1}-12)^2-4(1+m_1^2)(20)=0

$$( - 4 m _ { 1 } - 12 ) ^ { 2 } - 4 ( 1 + m _ { 1 } ^ { 2 } ) ( 20 ) = 0$$

Solve for m_1

$$m_{1}=2$$

$$m_{1}=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4m_{1}-12\right)^{2}.

$$16m_{1}^{2}+96m_{1}+144-4\left(1+m_{1}^{2}\right)\times 20=0$$

Multiply 4 and 20 to get 80.

$$16m_{1}^{2}+96m_{1}+144-80\left(1+m_{1}^{2}\right)=0$$

Use the distributive property to multiply -80 by 1+m_{1}^{2}.

$$16m_{1}^{2}+96m_{1}+144-80-80m_{1}^{2}=0$$

Subtract 80 from 144 to get 64.

$$16m_{1}^{2}+96m_{1}+64-80m_{1}^{2}=0$$

Combine 16m_{1}^{2} and -80m_{1}^{2} to get -64m_{1}^{2}.

$$-64m_{1}^{2}+96m_{1}+64=0$$

Divide both sides by 32.

$$-2m_{1}^{2}+3m_{1}+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2m_{1}^{2}+am_{1}+bm_{1}+2. To find a and b, set up a system to be solved.

$$a+b=3 ab=-2\times 2=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -2m_{1}^{2}+3m_{1}+2 as \left(-2m_{1}^{2}+4m_{1}\right)+\left(-m_{1}+2\right).

$$\left(-2m_{1}^{2}+4m_{1}\right)+\left(-m_{1}+2\right)$$

Factor out 2m_{1} in -2m_{1}^{2}+4m_{1}.

$$2m_{1}\left(-m_{1}+2\right)-m_{1}+2$$

Factor out common term -m_{1}+2 by using distributive property.

$$\left(-m_{1}+2\right)\left(2m_{1}+1\right)$$

To find equation solutions, solve -m_{1}+2=0 and 2m_{1}+1=0.

$$m_{1}=2 m_{1}=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4m_{1}-12\right)^{2}.

$$16m_{1}^{2}+96m_{1}+144-4\left(1+m_{1}^{2}\right)\times 20=0$$

Multiply 4 and 20 to get 80.

$$16m_{1}^{2}+96m_{1}+144-80\left(1+m_{1}^{2}\right)=0$$

Use the distributive property to multiply -80 by 1+m_{1}^{2}.

$$16m_{1}^{2}+96m_{1}+144-80-80m_{1}^{2}=0$$

Subtract 80 from 144 to get 64.

$$16m_{1}^{2}+96m_{1}+64-80m_{1}^{2}=0$$

Combine 16m_{1}^{2} and -80m_{1}^{2} to get -64m_{1}^{2}.

$$-64m_{1}^{2}+96m_{1}+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -64 for a, 96 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m_{1}=\frac{-96±\sqrt{96^{2}-4\left(-64\right)\times 64}}{2\left(-64\right)}$$

Square 96.

$$m_{1}=\frac{-96±\sqrt{9216-4\left(-64\right)\times 64}}{2\left(-64\right)}$$

Multiply -4 times -64.

$$m_{1}=\frac{-96±\sqrt{9216+256\times 64}}{2\left(-64\right)}$$

Multiply 256 times 64.

$$m_{1}=\frac{-96±\sqrt{9216+16384}}{2\left(-64\right)}$$

Add 9216 to 16384.

$$m_{1}=\frac{-96±\sqrt{25600}}{2\left(-64\right)}$$

Take the square root of 25600.

$$m_{1}=\frac{-96±160}{2\left(-64\right)}$$

Multiply 2 times -64.

$$m_{1}=\frac{-96±160}{-128}$$

Now solve the equation m_{1}=\frac{-96±160}{-128} when ± is plus. Add -96 to 160.

$$m_{1}=\frac{64}{-128}$$

Reduce the fraction \frac{64}{-128} to lowest terms by extracting and canceling out 64.

$$m_{1}=-\frac{1}{2}$$

Now solve the equation m_{1}=\frac{-96±160}{-128} when ± is minus. Subtract 160 from -96.

$$m_{1}=-\frac{256}{-128}$$

Divide -256 by -128.

$$m_{1}=2$$

The equation is now solved.

$$m_{1}=-\frac{1}{2} m_{1}=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4m_{1}-12\right)^{2}.

$$16m_{1}^{2}+96m_{1}+144-4\left(1+m_{1}^{2}\right)\times 20=0$$

Multiply 4 and 20 to get 80.

$$16m_{1}^{2}+96m_{1}+144-80\left(1+m_{1}^{2}\right)=0$$

Use the distributive property to multiply -80 by 1+m_{1}^{2}.

$$16m_{1}^{2}+96m_{1}+144-80-80m_{1}^{2}=0$$

Subtract 80 from 144 to get 64.

$$16m_{1}^{2}+96m_{1}+64-80m_{1}^{2}=0$$

Combine 16m_{1}^{2} and -80m_{1}^{2} to get -64m_{1}^{2}.

$$-64m_{1}^{2}+96m_{1}+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$-64m_{1}^{2}+96m_{1}=-64$$

Divide both sides by -64.

$$\frac{-64m_{1}^{2}+96m_{1}}{-64}=-\frac{64}{-64}$$

Dividing by -64 undoes the multiplication by -64.

$$m_{1}^{2}+\frac{96}{-64}m_{1}=-\frac{64}{-64}$$

Reduce the fraction \frac{96}{-64} to lowest terms by extracting and canceling out 32.

$$m_{1}^{2}-\frac{3}{2}m_{1}=-\frac{64}{-64}$$

Divide -64 by -64.

$$m_{1}^{2}-\frac{3}{2}m_{1}=1$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m_{1}^{2}-\frac{3}{2}m_{1}+\left(-\frac{3}{4}\right)^{2}=1+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$m_{1}^{2}-\frac{3}{2}m_{1}+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$m_{1}^{2}-\frac{3}{2}m_{1}+\frac{9}{16}=\frac{25}{16}$$

Factor m_{1}^{2}-\frac{3}{2}m_{1}+\frac{9}{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}.

$$\left(m_{1}-\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m_{1}-\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$m_{1}-\frac{3}{4}=\frac{5}{4} m_{1}-\frac{3}{4}=-\frac{5}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$m_{1}=2 m_{1}=-\frac{1}{2}$$