Solve 4t^2/5=-3t/5+9/10

$$\frac { 4 t ^ { 2 } } { 5 } = \frac { - 3 t } { 5 } + \frac { 9 } { 10 }$$

Solve for t

$$t = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$t=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 10, the least common multiple of 5,10.

$$2\times 4t^{2}=2\left(-3\right)t+9$$

Multiply 2 and 4 to get 8.

$$8t^{2}=2\left(-3\right)t+9$$

Multiply 2 and -3 to get -6.

$$8t^{2}=-6t+9$$

Add 6t to both sides.

$$8t^{2}+6t=9$$

Subtract 9 from both sides.

$$8t^{2}+6t-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8t^{2}+at+bt-9. To find a and b, set up a system to be solved.

$$a+b=6 ab=8\left(-9\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=-6 b=12$$

Rewrite 8t^{2}+6t-9 as \left(8t^{2}-6t\right)+\left(12t-9\right).

$$\left(8t^{2}-6t\right)+\left(12t-9\right)$$

Factor out 2t in the first and 3 in the second group.

$$2t\left(4t-3\right)+3\left(4t-3\right)$$

Factor out common term 4t-3 by using distributive property.

$$\left(4t-3\right)\left(2t+3\right)$$

To find equation solutions, solve 4t-3=0 and 2t+3=0.

$$t=\frac{3}{4} t=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 10, the least common multiple of 5,10.

$$2\times 4t^{2}=2\left(-3\right)t+9$$

Multiply 2 and 4 to get 8.

$$8t^{2}=2\left(-3\right)t+9$$

Multiply 2 and -3 to get -6.

$$8t^{2}=-6t+9$$

Add 6t to both sides.

$$8t^{2}+6t=9$$

Subtract 9 from both sides.

$$8t^{2}+6t-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-6±\sqrt{6^{2}-4\times 8\left(-9\right)}}{2\times 8}$$

Square 6.

$$t=\frac{-6±\sqrt{36-4\times 8\left(-9\right)}}{2\times 8}$$

Multiply -4 times 8.

$$t=\frac{-6±\sqrt{36-32\left(-9\right)}}{2\times 8}$$

Multiply -32 times -9.

$$t=\frac{-6±\sqrt{36+288}}{2\times 8}$$

Add 36 to 288.

$$t=\frac{-6±\sqrt{324}}{2\times 8}$$

Take the square root of 324.

$$t=\frac{-6±18}{2\times 8}$$

Multiply 2 times 8.

$$t=\frac{-6±18}{16}$$

Now solve the equation t=\frac{-6±18}{16} when ± is plus. Add -6 to 18.

$$t=\frac{12}{16}$$

Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.

$$t=\frac{3}{4}$$

Now solve the equation t=\frac{-6±18}{16} when ± is minus. Subtract 18 from -6.

$$t=-\frac{24}{16}$$

Reduce the fraction \frac{-24}{16} to lowest terms by extracting and canceling out 8.

$$t=-\frac{3}{2}$$

The equation is now solved.

$$t=\frac{3}{4} t=-\frac{3}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 10, the least common multiple of 5,10.

$$2\times 4t^{2}=2\left(-3\right)t+9$$

Multiply 2 and 4 to get 8.

$$8t^{2}=2\left(-3\right)t+9$$

Multiply 2 and -3 to get -6.

$$8t^{2}=-6t+9$$

Add 6t to both sides.

$$8t^{2}+6t=9$$

Divide both sides by 8.

$$\frac{8t^{2}+6t}{8}=\frac{9}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$t^{2}+\frac{6}{8}t=\frac{9}{8}$$

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

$$t^{2}+\frac{3}{4}t=\frac{9}{8}$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{3}{4}t+\left(\frac{3}{8}\right)^{2}=\frac{9}{8}+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{3}{4}t+\frac{9}{64}=\frac{9}{8}+\frac{9}{64}$$

Add \frac{9}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{3}{4}t+\frac{9}{64}=\frac{81}{64}$$

Factor t^{2}+\frac{3}{4}t+\frac{9}{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}.

$$\left(t+\frac{3}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{3}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$t+\frac{3}{8}=\frac{9}{8} t+\frac{3}{8}=-\frac{9}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$t=\frac{3}{4} t=-\frac{3}{2}$$