Solve -2/5(4-5a)^2+7/5(4-5a)+24/5=4a-1

$$- \frac { 2 } { 5 } ( 4 - 5 a ) ^ { 2 } + \frac { 7 } { 5 } ( 4 - 5 a ) + \frac { 24 } { 5 } = 4 a - 1$$

Solve for a

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

$$a=1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-5a\right)^{2}.

$$-\frac{2}{5}\left(16-40a+25a^{2}\right)+\frac{7}{5}\left(4-5a\right)+\frac{24}{5}=4a-1$$

Use the distributive property to multiply -\frac{2}{5} by 16-40a+25a^{2}.

$$-\frac{32}{5}+16a-10a^{2}+\frac{7}{5}\left(4-5a\right)+\frac{24}{5}=4a-1$$

Use the distributive property to multiply \frac{7}{5} by 4-5a.

$$-\frac{32}{5}+16a-10a^{2}+\frac{28}{5}-7a+\frac{24}{5}=4a-1$$

Add -\frac{32}{5} and \frac{28}{5} to get -\frac{4}{5}.

$$-\frac{4}{5}+16a-10a^{2}-7a+\frac{24}{5}=4a-1$$

Combine 16a and -7a to get 9a.

$$-\frac{4}{5}+9a-10a^{2}+\frac{24}{5}=4a-1$$

Add -\frac{4}{5} and \frac{24}{5} to get 4.

$$4+9a-10a^{2}=4a-1$$

Subtract 4a from both sides.

$$4+9a-10a^{2}-4a=-1$$

Combine 9a and -4a to get 5a.

$$4+5a-10a^{2}=-1$$

Add 1 to both sides.

$$4+5a-10a^{2}+1=0$$

Add 4 and 1 to get 5.

$$5+5a-10a^{2}=0$$

Divide both sides by 5.

$$1+a-2a^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2a^{2}+a+1=0$$

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

$$a+b=1 ab=-2=-2$$

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. The only such pair is the system solution.

$$a=2 b=-1$$

Rewrite -2a^{2}+a+1 as \left(-2a^{2}+2a\right)+\left(-a+1\right).

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

Factor out 2a in -2a^{2}+2a.

$$2a\left(-a+1\right)-a+1$$

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

$$\left(-a+1\right)\left(2a+1\right)$$

To find equation solutions, solve -a+1=0 and 2a+1=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-5a\right)^{2}.

$$-\frac{2}{5}\left(16-40a+25a^{2}\right)+\frac{7}{5}\left(4-5a\right)+\frac{24}{5}=4a-1$$

Use the distributive property to multiply -\frac{2}{5} by 16-40a+25a^{2}.

$$-\frac{32}{5}+16a-10a^{2}+\frac{7}{5}\left(4-5a\right)+\frac{24}{5}=4a-1$$

Use the distributive property to multiply \frac{7}{5} by 4-5a.

$$-\frac{32}{5}+16a-10a^{2}+\frac{28}{5}-7a+\frac{24}{5}=4a-1$$

Add -\frac{32}{5} and \frac{28}{5} to get -\frac{4}{5}.

$$-\frac{4}{5}+16a-10a^{2}-7a+\frac{24}{5}=4a-1$$

Combine 16a and -7a to get 9a.

$$-\frac{4}{5}+9a-10a^{2}+\frac{24}{5}=4a-1$$

Add -\frac{4}{5} and \frac{24}{5} to get 4.

$$4+9a-10a^{2}=4a-1$$

Subtract 4a from both sides.

$$4+9a-10a^{2}-4a=-1$$

Combine 9a and -4a to get 5a.

$$4+5a-10a^{2}=-1$$

Add 1 to both sides.

$$4+5a-10a^{2}+1=0$$

Add 4 and 1 to get 5.

$$5+5a-10a^{2}=0$$

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.

$$-10a^{2}+5a+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 5 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-5±\sqrt{5^{2}-4\left(-10\right)\times 5}}{2\left(-10\right)}$$

Square 5.

$$a=\frac{-5±\sqrt{25-4\left(-10\right)\times 5}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$a=\frac{-5±\sqrt{25+40\times 5}}{2\left(-10\right)}$$

Multiply 40 times 5.

$$a=\frac{-5±\sqrt{25+200}}{2\left(-10\right)}$$

Add 25 to 200.

$$a=\frac{-5±\sqrt{225}}{2\left(-10\right)}$$

Take the square root of 225.

$$a=\frac{-5±15}{2\left(-10\right)}$$

Multiply 2 times -10.

$$a=\frac{-5±15}{-20}$$

Now solve the equation a=\frac{-5±15}{-20} when ± is plus. Add -5 to 15.

$$a=\frac{10}{-20}$$

Reduce the fraction \frac{10}{-20} to lowest terms by extracting and canceling out 10.

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

Now solve the equation a=\frac{-5±15}{-20} when ± is minus. Subtract 15 from -5.

$$a=-\frac{20}{-20}$$

Divide -20 by -20.

$$a=1$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-5a\right)^{2}.

$$-\frac{2}{5}\left(16-40a+25a^{2}\right)+\frac{7}{5}\left(4-5a\right)+\frac{24}{5}=4a-1$$

Use the distributive property to multiply -\frac{2}{5} by 16-40a+25a^{2}.

$$-\frac{32}{5}+16a-10a^{2}+\frac{7}{5}\left(4-5a\right)+\frac{24}{5}=4a-1$$

Use the distributive property to multiply \frac{7}{5} by 4-5a.

$$-\frac{32}{5}+16a-10a^{2}+\frac{28}{5}-7a+\frac{24}{5}=4a-1$$

Add -\frac{32}{5} and \frac{28}{5} to get -\frac{4}{5}.

$$-\frac{4}{5}+16a-10a^{2}-7a+\frac{24}{5}=4a-1$$

Combine 16a and -7a to get 9a.

$$-\frac{4}{5}+9a-10a^{2}+\frac{24}{5}=4a-1$$

Add -\frac{4}{5} and \frac{24}{5} to get 4.

$$4+9a-10a^{2}=4a-1$$

Subtract 4a from both sides.

$$4+9a-10a^{2}-4a=-1$$

Combine 9a and -4a to get 5a.

$$4+5a-10a^{2}=-1$$

Subtract 4 from both sides.

$$5a-10a^{2}=-1-4$$

Subtract 4 from -1 to get -5.

$$5a-10a^{2}=-5$$

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.

$$-10a^{2}+5a=-5$$

Divide both sides by -10.

$$\frac{-10a^{2}+5a}{-10}=-\frac{5}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$a^{2}+\frac{5}{-10}a=-\frac{5}{-10}$$

Reduce the fraction \frac{5}{-10} to lowest terms by extracting and canceling out 5.

$$a^{2}-\frac{1}{2}a=-\frac{5}{-10}$$

Reduce the fraction \frac{-5}{-10} to lowest terms by extracting and canceling out 5.

$$a^{2}-\frac{1}{2}a=\frac{1}{2}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor a^{2}-\frac{1}{2}a+\frac{1}{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}.

$$\left(a-\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$a-\frac{1}{4}=\frac{3}{4} a-\frac{1}{4}=-\frac{3}{4}$$

Add \frac{1}{4} to both sides of the equation.

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