Solve -5t^2-3t+2

$$- 5 t ^ { 2 } - 3 t + 2$$

Factor

$$-\left(5t-2\right)\left(t+1\right)$$

Solution Steps

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

$$a+b=-3 ab=-5\times 2=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=2 b=-5$$

Rewrite -5t^{2}-3t+2 as \left(-5t^{2}+2t\right)+\left(-5t+2\right).

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

Factor out -t in the first and -1 in the second group.

$$-t\left(5t-2\right)-\left(5t-2\right)$$

Factor out common term 5t-2 by using distributive property.

$$\left(5t-2\right)\left(-t-1\right)$$

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.

$$-5t^{2}-3t+2=0$$

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.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-5\right)\times 2}}{2\left(-5\right)}$$

Square -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-4\left(-5\right)\times 2}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-\left(-3\right)±\sqrt{9+20\times 2}}{2\left(-5\right)}$$

Multiply 20 times 2.

$$t=\frac{-\left(-3\right)±\sqrt{9+40}}{2\left(-5\right)}$$

Add 9 to 40.

$$t=\frac{-\left(-3\right)±\sqrt{49}}{2\left(-5\right)}$$

Take the square root of 49.

$$t=\frac{-\left(-3\right)±7}{2\left(-5\right)}$$

The opposite of -3 is 3.

$$t=\frac{3±7}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{3±7}{-10}$$

Now solve the equation t=\frac{3±7}{-10} when ± is plus. Add 3 to 7.

$$t=\frac{10}{-10}$$

Divide 10 by -10.

$$t=-1$$

Now solve the equation t=\frac{3±7}{-10} when ± is minus. Subtract 7 from 3.

$$t=-\frac{4}{-10}$$

Reduce the fraction \frac{-4}{-10} to lowest terms by extracting and canceling out 2.

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

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

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

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

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

Subtract \frac{2}{5} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 5, the greatest common factor in -5 and 5.

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

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.

$$x ^ 2 +\frac{3}{5}x -\frac{2}{5} = 0$$

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

$$r + s = -\frac{3}{5} rs = -\frac{2}{5}$$

Two numbers r and s sum up to -\frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{5} = -\frac{3}{10}. 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>

$$r = -\frac{3}{10} - u s = -\frac{3}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{5}

$$(-\frac{3}{10} - u) (-\frac{3}{10} + u) = -\frac{2}{5}$$

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

$$\frac{9}{100} - u^2 = -\frac{2}{5}$$

Simplify the expression by subtracting \frac{9}{100} on both sides

$$-u^2 = -\frac{2}{5}-\frac{9}{100} = -\frac{49}{100}$$

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

$$u^2 = \frac{49}{100} u = \pm\sqrt{\frac{49}{100}} = \pm \frac{7}{10}$$

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

$$r =-\frac{3}{10} - \frac{7}{10} = -1 s = -\frac{3}{10} + \frac{7}{10} = 0.400$$

Evaluate

$$-\left(5t-2\right)\left(t+1\right)$$