Solve (-700-42k)^2-4(490)(300+k^2+20k)=0

$$( - 700 - 42 k ) ^ { 2 } - 4 ( 490 ) ( 300 + k ^ { 2 } + 20 k ) = 0$$

Solve for k

$$k=20\sqrt{5}+50\approx 94.72135955$$

$$k=50-20\sqrt{5}\approx 5.27864045$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-700-42k\right)^{2}.

$$490000+58800k+1764k^{2}-4\times 490\left(300+k^{2}+20k\right)=0$$

Multiply 4 and 490 to get 1960.

$$490000+58800k+1764k^{2}-1960\left(300+k^{2}+20k\right)=0$$

Use the distributive property to multiply -1960 by 300+k^{2}+20k.

$$490000+58800k+1764k^{2}-588000-1960k^{2}-39200k=0$$

Subtract 588000 from 490000 to get -98000.

$$-98000+58800k+1764k^{2}-1960k^{2}-39200k=0$$

Combine 1764k^{2} and -1960k^{2} to get -196k^{2}.

$$-98000+58800k-196k^{2}-39200k=0$$

Combine 58800k and -39200k to get 19600k.

$$-98000+19600k-196k^{2}=0$$

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.

$$-196k^{2}+19600k-98000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -196 for a, 19600 for b, and -98000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-19600±\sqrt{19600^{2}-4\left(-196\right)\left(-98000\right)}}{2\left(-196\right)}$$

Square 19600.

$$k=\frac{-19600±\sqrt{384160000-4\left(-196\right)\left(-98000\right)}}{2\left(-196\right)}$$

Multiply -4 times -196.

$$k=\frac{-19600±\sqrt{384160000+784\left(-98000\right)}}{2\left(-196\right)}$$

Multiply 784 times -98000.

$$k=\frac{-19600±\sqrt{384160000-76832000}}{2\left(-196\right)}$$

Add 384160000 to -76832000.

$$k=\frac{-19600±\sqrt{307328000}}{2\left(-196\right)}$$

Take the square root of 307328000.

$$k=\frac{-19600±7840\sqrt{5}}{2\left(-196\right)}$$

Multiply 2 times -196.

$$k=\frac{-19600±7840\sqrt{5}}{-392}$$

Now solve the equation k=\frac{-19600±7840\sqrt{5}}{-392} when ± is plus. Add -19600 to 7840\sqrt{5}.

$$k=\frac{7840\sqrt{5}-19600}{-392}$$

Divide -19600+7840\sqrt{5} by -392.

$$k=50-20\sqrt{5}$$

Now solve the equation k=\frac{-19600±7840\sqrt{5}}{-392} when ± is minus. Subtract 7840\sqrt{5} from -19600.

$$k=\frac{-7840\sqrt{5}-19600}{-392}$$

Divide -19600-7840\sqrt{5} by -392.

$$k=20\sqrt{5}+50$$

The equation is now solved.

$$k=50-20\sqrt{5} k=20\sqrt{5}+50$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-700-42k\right)^{2}.

$$490000+58800k+1764k^{2}-4\times 490\left(300+k^{2}+20k\right)=0$$

Multiply 4 and 490 to get 1960.

$$490000+58800k+1764k^{2}-1960\left(300+k^{2}+20k\right)=0$$

Use the distributive property to multiply -1960 by 300+k^{2}+20k.

$$490000+58800k+1764k^{2}-588000-1960k^{2}-39200k=0$$

Subtract 588000 from 490000 to get -98000.

$$-98000+58800k+1764k^{2}-1960k^{2}-39200k=0$$

Combine 1764k^{2} and -1960k^{2} to get -196k^{2}.

$$-98000+58800k-196k^{2}-39200k=0$$

Combine 58800k and -39200k to get 19600k.

$$-98000+19600k-196k^{2}=0$$

Add 98000 to both sides. Anything plus zero gives itself.

$$19600k-196k^{2}=98000$$

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.

$$-196k^{2}+19600k=98000$$

Divide both sides by -196.

$$\frac{-196k^{2}+19600k}{-196}=\frac{98000}{-196}$$

Dividing by -196 undoes the multiplication by -196.

$$k^{2}+\frac{19600}{-196}k=\frac{98000}{-196}$$

Divide 19600 by -196.

$$k^{2}-100k=\frac{98000}{-196}$$

Divide 98000 by -196.

$$k^{2}-100k=-500$$

Divide -100, the coefficient of the x term, by 2 to get -50. Then add the square of -50 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-100k+\left(-50\right)^{2}=-500+\left(-50\right)^{2}$$

Square -50.

$$k^{2}-100k+2500=-500+2500$$

Add -500 to 2500.

$$k^{2}-100k+2500=2000$$

Factor k^{2}-100k+2500. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-50\right)^{2}=2000$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-50\right)^{2}}=\sqrt{2000}$$

Simplify.

$$k-50=20\sqrt{5} k-50=-20\sqrt{5}$$

Add 50 to both sides of the equation.

$$k=20\sqrt{5}+50 k=50-20\sqrt{5}$$