Solve 18t^2-9t-5

$$18 t ^ { 2 } - 9 t - 5$$

Factor

$$\left(6t-5\right)\left(3t+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 18t^{2}+at+bt-5. To find a and b, set up a system to be solved.

$$a+b=-9 ab=18\left(-5\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -9.

$$a=-15 b=6$$

Rewrite 18t^{2}-9t-5 as \left(18t^{2}-15t\right)+\left(6t-5\right).

$$\left(18t^{2}-15t\right)+\left(6t-5\right)$$

Factor out 3t in 18t^{2}-15t.

$$3t\left(6t-5\right)+6t-5$$

Factor out common term 6t-5 by using distributive property.

$$\left(6t-5\right)\left(3t+1\right)$$

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.

$$18t^{2}-9t-5=0$$

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.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 18\left(-5\right)}}{2\times 18}$$

Square -9.

$$t=\frac{-\left(-9\right)±\sqrt{81-4\times 18\left(-5\right)}}{2\times 18}$$

Multiply -4 times 18.

$$t=\frac{-\left(-9\right)±\sqrt{81-72\left(-5\right)}}{2\times 18}$$

Multiply -72 times -5.

$$t=\frac{-\left(-9\right)±\sqrt{81+360}}{2\times 18}$$

Add 81 to 360.

$$t=\frac{-\left(-9\right)±\sqrt{441}}{2\times 18}$$

Take the square root of 441.

$$t=\frac{-\left(-9\right)±21}{2\times 18}$$

The opposite of -9 is 9.

$$t=\frac{9±21}{2\times 18}$$

Multiply 2 times 18.

$$t=\frac{9±21}{36}$$

Now solve the equation t=\frac{9±21}{36} when ± is plus. Add 9 to 21.

$$t=\frac{30}{36}$$

Reduce the fraction \frac{30}{36} to lowest terms by extracting and canceling out 6.

$$t=\frac{5}{6}$$

Now solve the equation t=\frac{9±21}{36} when ± is minus. Subtract 21 from 9.

$$t=-\frac{12}{36}$$

Reduce the fraction \frac{-12}{36} to lowest terms by extracting and canceling out 12.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{6} for x_{1} and -\frac{1}{3} for x_{2}.

$$18t^{2}-9t-5=18\left(t-\frac{5}{6}\right)\left(t-\left(-\frac{1}{3}\right)\right)$$

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

$$18t^{2}-9t-5=18\left(t-\frac{5}{6}\right)\left(t+\frac{1}{3}\right)$$

Subtract \frac{5}{6} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18t^{2}-9t-5=18\times \frac{6t-5}{6}\left(t+\frac{1}{3}\right)$$

Add \frac{1}{3} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18t^{2}-9t-5=18\times \frac{6t-5}{6}\times \frac{3t+1}{3}$$

Multiply \frac{6t-5}{6} times \frac{3t+1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18t^{2}-9t-5=18\times \frac{\left(6t-5\right)\left(3t+1\right)}{6\times 3}$$

Multiply 6 times 3.

$$18t^{2}-9t-5=18\times \frac{\left(6t-5\right)\left(3t+1\right)}{18}$$

Cancel out 18, the greatest common factor in 18 and 18.

$$18t^{2}-9t-5=\left(6t-5\right)\left(3t+1\right)$$

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 18

$$x ^ 2 -\frac{1}{2}x -\frac{5}{18} = 0$$

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

$$r + s = \frac{1}{2} rs = -\frac{5}{18}$$

Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. 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>

$$r = \frac{1}{4} - u s = \frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{18}

$$(\frac{1}{4} - u) (\frac{1}{4} + u) = -\frac{5}{18}$$

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

$$\frac{1}{16} - u^2 = -\frac{5}{18}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{5}{18}-\frac{1}{16} = -\frac{49}{144}$$

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

$$u^2 = \frac{49}{144} u = \pm\sqrt{\frac{49}{144}} = \pm \frac{7}{12}$$

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

$$r =\frac{1}{4} - \frac{7}{12} = -0.333 s = \frac{1}{4} + \frac{7}{12} = 0.833$$

Evaluate

$$\left(6t-5\right)\left(3t+1\right)$$