Solve t^2+3t-10=4+8t?

$$t ^ { 2 } + 3 t - 10 = 4 + 8 t ?$$

Solve for t

$$t=-2$$

$$t=7$$

Steps Using Factoring

Subtract 4 from both sides.

$$t^{2}+3t-10-4=8t$$

Subtract 4 from -10 to get -14.

$$t^{2}+3t-14=8t$$

Subtract 8t from both sides.

$$t^{2}+3t-14-8t=0$$

Combine 3t and -8t to get -5t.

$$t^{2}-5t-14=0$$

To solve the equation, factor t^{2}-5t-14 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=-14$$

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 -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=-7 b=2$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t-7\right)\left(t+2\right)$$

To find equation solutions, solve t-7=0 and t+2=0.

$$t=7 t=-2$$

Steps Using Factoring By Grouping

Subtract 4 from both sides.

$$t^{2}+3t-10-4=8t$$

Subtract 4 from -10 to get -14.

$$t^{2}+3t-14=8t$$

Subtract 8t from both sides.

$$t^{2}+3t-14-8t=0$$

Combine 3t and -8t to get -5t.

$$t^{2}-5t-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-14. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-14\right)=-14$$

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 -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=-7 b=2$$

Rewrite t^{2}-5t-14 as \left(t^{2}-7t\right)+\left(2t-14\right).

$$\left(t^{2}-7t\right)+\left(2t-14\right)$$

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

$$t\left(t-7\right)+2\left(t-7\right)$$

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

$$\left(t-7\right)\left(t+2\right)$$

To find equation solutions, solve t-7=0 and t+2=0.

$$t=7 t=-2$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$t^{2}+3t-10-4=8t$$

Subtract 4 from -10 to get -14.

$$t^{2}+3t-14=8t$$

Subtract 8t from both sides.

$$t^{2}+3t-14-8t=0$$

Combine 3t and -8t to get -5t.

$$t^{2}-5t-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-14\right)}}{2}$$

Square -5.

$$t=\frac{-\left(-5\right)±\sqrt{25-4\left(-14\right)}}{2}$$

Multiply -4 times -14.

$$t=\frac{-\left(-5\right)±\sqrt{25+56}}{2}$$

Add 25 to 56.

$$t=\frac{-\left(-5\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$t=\frac{-\left(-5\right)±9}{2}$$

The opposite of -5 is 5.

$$t=\frac{5±9}{2}$$

Now solve the equation t=\frac{5±9}{2} when ± is plus. Add 5 to 9.

$$t=\frac{14}{2}$$

Divide 14 by 2.

$$t=7$$

Now solve the equation t=\frac{5±9}{2} when ± is minus. Subtract 9 from 5.

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

Divide -4 by 2.

$$t=-2$$

The equation is now solved.

$$t=7 t=-2$$

Steps for Completing the Square

Subtract 8t from both sides.

$$t^{2}+3t-10-8t=4$$

Combine 3t and -8t to get -5t.

$$t^{2}-5t-10=4$$

Add 10 to both sides.

$$t^{2}-5t=4+10$$

Add 4 and 10 to get 14.

$$t^{2}-5t=14$$

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

$$t^{2}-5t+\left(-\frac{5}{2}\right)^{2}=14+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-5t+\frac{25}{4}=14+\frac{25}{4}$$

Add 14 to \frac{25}{4}.

$$t^{2}-5t+\frac{25}{4}=\frac{81}{4}$$

Factor t^{2}-5t+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{5}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$t-\frac{5}{2}=\frac{9}{2} t-\frac{5}{2}=-\frac{9}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$t=7 t=-2$$