Solve (-(6c+10))^2-4(10)(c^2+2c-5)=0

$$( - ( 6 c + 10 ) ) ^ { 2 } - 4 ( 10 ) ( c ^ { 2 } + 2 c - 5 ) = 0$$

Solve for c

$$c=-5$$

$$c=15$$

Steps Using Factoring By Grouping

To find the opposite of 6c+10, find the opposite of each term.

$$\left(-6c-10\right)^{2}-4\times 10\left(c^{2}+2c-5\right)=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6c-10\right)^{2}.

$$36c^{2}+120c+100-4\times 10\left(c^{2}+2c-5\right)=0$$

Multiply 4 and 10 to get 40.

$$36c^{2}+120c+100-40\left(c^{2}+2c-5\right)=0$$

Use the distributive property to multiply -40 by c^{2}+2c-5.

$$36c^{2}+120c+100-40c^{2}-80c+200=0$$

Combine 36c^{2} and -40c^{2} to get -4c^{2}.

$$-4c^{2}+120c+100-80c+200=0$$

Combine 120c and -80c to get 40c.

$$-4c^{2}+40c+100+200=0$$

Add 100 and 200 to get 300.

$$-4c^{2}+40c+300=0$$

Divide both sides by 4.

$$-c^{2}+10c+75=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -c^{2}+ac+bc+75. To find a and b, set up a system to be solved.

$$a+b=10 ab=-75=-75$$

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 -75.

$$-1,75 -3,25 -5,15$$

Calculate the sum for each pair.

$$-1+75=74 -3+25=22 -5+15=10$$

The solution is the pair that gives sum 10.

$$a=15 b=-5$$

Rewrite -c^{2}+10c+75 as \left(-c^{2}+15c\right)+\left(-5c+75\right).

$$\left(-c^{2}+15c\right)+\left(-5c+75\right)$$

Factor out -c in the first and -5 in the second group.

$$-c\left(c-15\right)-5\left(c-15\right)$$

Factor out common term c-15 by using distributive property.

$$\left(c-15\right)\left(-c-5\right)$$

To find equation solutions, solve c-15=0 and -c-5=0.

$$c=15 c=-5$$

Steps Using the Quadratic Formula

To find the opposite of 6c+10, find the opposite of each term.

$$\left(-6c-10\right)^{2}-4\times 10\left(c^{2}+2c-5\right)=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6c-10\right)^{2}.

$$36c^{2}+120c+100-4\times 10\left(c^{2}+2c-5\right)=0$$

Multiply 4 and 10 to get 40.

$$36c^{2}+120c+100-40\left(c^{2}+2c-5\right)=0$$

Use the distributive property to multiply -40 by c^{2}+2c-5.

$$36c^{2}+120c+100-40c^{2}-80c+200=0$$

Combine 36c^{2} and -40c^{2} to get -4c^{2}.

$$-4c^{2}+120c+100-80c+200=0$$

Combine 120c and -80c to get 40c.

$$-4c^{2}+40c+100+200=0$$

Add 100 and 200 to get 300.

$$-4c^{2}+40c+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 40 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-40±\sqrt{40^{2}-4\left(-4\right)\times 300}}{2\left(-4\right)}$$

Square 40.

$$c=\frac{-40±\sqrt{1600-4\left(-4\right)\times 300}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$c=\frac{-40±\sqrt{1600+16\times 300}}{2\left(-4\right)}$$

Multiply 16 times 300.

$$c=\frac{-40±\sqrt{1600+4800}}{2\left(-4\right)}$$

Add 1600 to 4800.

$$c=\frac{-40±\sqrt{6400}}{2\left(-4\right)}$$

Take the square root of 6400.

$$c=\frac{-40±80}{2\left(-4\right)}$$

Multiply 2 times -4.

$$c=\frac{-40±80}{-8}$$

Now solve the equation c=\frac{-40±80}{-8} when ± is plus. Add -40 to 80.

$$c=\frac{40}{-8}$$

Divide 40 by -8.

$$c=-5$$

Now solve the equation c=\frac{-40±80}{-8} when ± is minus. Subtract 80 from -40.

$$c=-\frac{120}{-8}$$

Divide -120 by -8.

$$c=15$$

The equation is now solved.

$$c=-5 c=15$$

Steps for Completing the Square

To find the opposite of 6c+10, find the opposite of each term.

$$\left(-6c-10\right)^{2}-4\times 10\left(c^{2}+2c-5\right)=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6c-10\right)^{2}.

$$36c^{2}+120c+100-4\times 10\left(c^{2}+2c-5\right)=0$$

Multiply 4 and 10 to get 40.

$$36c^{2}+120c+100-40\left(c^{2}+2c-5\right)=0$$

Use the distributive property to multiply -40 by c^{2}+2c-5.

$$36c^{2}+120c+100-40c^{2}-80c+200=0$$

Combine 36c^{2} and -40c^{2} to get -4c^{2}.

$$-4c^{2}+120c+100-80c+200=0$$

Combine 120c and -80c to get 40c.

$$-4c^{2}+40c+100+200=0$$

Add 100 and 200 to get 300.

$$-4c^{2}+40c+300=0$$

Subtract 300 from both sides. Anything subtracted from zero gives its negation.

$$-4c^{2}+40c=-300$$

Divide both sides by -4.

$$\frac{-4c^{2}+40c}{-4}=-\frac{300}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$c^{2}+\frac{40}{-4}c=-\frac{300}{-4}$$

Divide 40 by -4.

$$c^{2}-10c=-\frac{300}{-4}$$

Divide -300 by -4.

$$c^{2}-10c=75$$

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

$$c^{2}-10c+\left(-5\right)^{2}=75+\left(-5\right)^{2}$$

Square -5.

$$c^{2}-10c+25=75+25$$

Add 75 to 25.

$$c^{2}-10c+25=100$$

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

$$\left(c-5\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-5\right)^{2}}=\sqrt{100}$$

Simplify.

$$c-5=10 c-5=-10$$

Add 5 to both sides of the equation.

$$c=15 c=-5$$