Solve 12t^2-7t-10

$$12 t ^ { 2 } - 7 t - 10$$

Factor

$$\left(4t-5\right)\left(3t+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12t^{2}+at+bt-10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=12\left(-10\right)=-120$$

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 -120.

$$1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12$$

Calculate the sum for each pair.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

The solution is the pair that gives sum -7.

$$a=-15 b=8$$

Rewrite 12t^{2}-7t-10 as \left(12t^{2}-15t\right)+\left(8t-10\right).

$$\left(12t^{2}-15t\right)+\left(8t-10\right)$$

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

$$3t\left(4t-5\right)+2\left(4t-5\right)$$

Factor out common term 4t-5 by using distributive property.

$$\left(4t-5\right)\left(3t+2\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$12t^{2}-7t-10=0$$

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.

$$t=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12\left(-10\right)}}{2\times 12}$$

Square -7.

$$t=\frac{-\left(-7\right)±\sqrt{49-4\times 12\left(-10\right)}}{2\times 12}$$

Multiply -4 times 12.

$$t=\frac{-\left(-7\right)±\sqrt{49-48\left(-10\right)}}{2\times 12}$$

Multiply -48 times -10.

$$t=\frac{-\left(-7\right)±\sqrt{49+480}}{2\times 12}$$

Add 49 to 480.

$$t=\frac{-\left(-7\right)±\sqrt{529}}{2\times 12}$$

Take the square root of 529.

$$t=\frac{-\left(-7\right)±23}{2\times 12}$$

The opposite of -7 is 7.

$$t=\frac{7±23}{2\times 12}$$

Multiply 2 times 12.

$$t=\frac{7±23}{24}$$

Now solve the equation t=\frac{7±23}{24} when ± is plus. Add 7 to 23.

$$t=\frac{30}{24}$$

Reduce the fraction \frac{30}{24} to lowest terms by extracting and canceling out 6.

$$t=\frac{5}{4}$$

Now solve the equation t=\frac{7±23}{24} when ± is minus. Subtract 23 from 7.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and -\frac{2}{3} for x_{2}.

$$12t^{2}-7t-10=12\left(t-\frac{5}{4}\right)\left(t-\left(-\frac{2}{3}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$12t^{2}-7t-10=12\left(t-\frac{5}{4}\right)\left(t+\frac{2}{3}\right)$$

Subtract \frac{5}{4} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12t^{2}-7t-10=12\times \frac{4t-5}{4}\left(t+\frac{2}{3}\right)$$

Add \frac{2}{3} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12t^{2}-7t-10=12\times \frac{4t-5}{4}\times \frac{3t+2}{3}$$

Multiply \frac{4t-5}{4} times \frac{3t+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12t^{2}-7t-10=12\times \frac{\left(4t-5\right)\left(3t+2\right)}{4\times 3}$$

Multiply 4 times 3.

$$12t^{2}-7t-10=12\times \frac{\left(4t-5\right)\left(3t+2\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12t^{2}-7t-10=\left(4t-5\right)\left(3t+2\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 12

$$x ^ 2 -\frac{7}{12}x -\frac{5}{6} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{7}{12} rs = -\frac{5}{6}$$

Two numbers r and s sum up to \frac{7}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{12} = \frac{7}{24}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{7}{24} - u s = \frac{7}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{6}

$$(\frac{7}{24} - u) (\frac{7}{24} + u) = -\frac{5}{6}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{49}{576} - u^2 = -\frac{5}{6}$$

Simplify the expression by subtracting \frac{49}{576} on both sides

$$-u^2 = -\frac{5}{6}-\frac{49}{576} = -\frac{529}{576}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{529}{576} u = \pm\sqrt{\frac{529}{576}} = \pm \frac{23}{24}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{7}{24} - \frac{23}{24} = -0.667 s = \frac{7}{24} + \frac{23}{24} = 1.250$$

Evaluate

$$\left(4t-5\right)\left(3t+2\right)$$