Solve 4c^2=25c+21

$$4 c ^ { 2 } = 25 c + 21$$

Solve for c

$$c=-\frac{3}{4}=-0.75$$

$$c=7$$

Steps Using Factoring By Grouping

Subtract 25c from both sides.

$$4c^{2}-25c=21$$

Subtract 21 from both sides.

$$4c^{2}-25c-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4c^{2}+ac+bc-21. To find a and b, set up a system to be solved.

$$a+b=-25 ab=4\left(-21\right)=-84$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -84.

$$1,-84 2,-42 3,-28 4,-21 6,-14 7,-12$$

Calculate the sum for each pair.

$$1-84=-83 2-42=-40 3-28=-25 4-21=-17 6-14=-8 7-12=-5$$

The solution is the pair that gives sum -25.

$$a=-28 b=3$$

Rewrite 4c^{2}-25c-21 as \left(4c^{2}-28c\right)+\left(3c-21\right).

$$\left(4c^{2}-28c\right)+\left(3c-21\right)$$

Factor out 4c in the first and 3 in the second group.

$$4c\left(c-7\right)+3\left(c-7\right)$$

Factor out common term c-7 by using distributive property.

$$\left(c-7\right)\left(4c+3\right)$$

To find equation solutions, solve c-7=0 and 4c+3=0.

$$c=7 c=-\frac{3}{4}$$

Steps Using the Quadratic Formula

Subtract 25c from both sides.

$$4c^{2}-25c=21$$

Subtract 21 from both sides.

$$4c^{2}-25c-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -25 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 4\left(-21\right)}}{2\times 4}$$

Square -25.

$$c=\frac{-\left(-25\right)±\sqrt{625-4\times 4\left(-21\right)}}{2\times 4}$$

Multiply -4 times 4.

$$c=\frac{-\left(-25\right)±\sqrt{625-16\left(-21\right)}}{2\times 4}$$

Multiply -16 times -21.

$$c=\frac{-\left(-25\right)±\sqrt{625+336}}{2\times 4}$$

Add 625 to 336.

$$c=\frac{-\left(-25\right)±\sqrt{961}}{2\times 4}$$

Take the square root of 961.

$$c=\frac{-\left(-25\right)±31}{2\times 4}$$

The opposite of -25 is 25.

$$c=\frac{25±31}{2\times 4}$$

Multiply 2 times 4.

$$c=\frac{25±31}{8}$$

Now solve the equation c=\frac{25±31}{8} when ± is plus. Add 25 to 31.

$$c=\frac{56}{8}$$

Divide 56 by 8.

$$c=7$$

Now solve the equation c=\frac{25±31}{8} when ± is minus. Subtract 31 from 25.

$$c=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$c=-\frac{3}{4}$$

The equation is now solved.

$$c=7 c=-\frac{3}{4}$$

Steps for Completing the Square

Subtract 25c from both sides.

$$4c^{2}-25c=21$$

Divide both sides by 4.

$$\frac{4c^{2}-25c}{4}=\frac{21}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$c^{2}-\frac{25}{4}c=\frac{21}{4}$$

Divide -\frac{25}{4}, the coefficient of the x term, by 2 to get -\frac{25}{8}. Then add the square of -\frac{25}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-\frac{25}{4}c+\left(-\frac{25}{8}\right)^{2}=\frac{21}{4}+\left(-\frac{25}{8}\right)^{2}$$

Square -\frac{25}{8} by squaring both the numerator and the denominator of the fraction.

$$c^{2}-\frac{25}{4}c+\frac{625}{64}=\frac{21}{4}+\frac{625}{64}$$

Add \frac{21}{4} to \frac{625}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$c^{2}-\frac{25}{4}c+\frac{625}{64}=\frac{961}{64}$$

Factor c^{2}-\frac{25}{4}c+\frac{625}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c-\frac{25}{8}\right)^{2}=\frac{961}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-\frac{25}{8}\right)^{2}}=\sqrt{\frac{961}{64}}$$

Simplify.

$$c-\frac{25}{8}=\frac{31}{8} c-\frac{25}{8}=-\frac{31}{8}$$

Add \frac{25}{8} to both sides of the equation.

$$c=7 c=-\frac{3}{4}$$