Solve h(t)=-5t^2+14t+3

$$h ( t ) = - 5 t ^ { 2 } + 14 t + 3$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -5t^{2}+at+bt+3. To find a and b, set up a system to be solved.

$$a+b=14 ab=-5\times 3=-15$$

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 -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 14.

$$a=15 b=-1$$

Rewrite -5t^{2}+14t+3 as \left(-5t^{2}+15t\right)+\left(-t+3\right).

$$\left(-5t^{2}+15t\right)+\left(-t+3\right)$$

Factor out 5t in -5t^{2}+15t.

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

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

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

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.

$$-5t^{2}+14t+3=0$$

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.

$$t=\frac{-14±\sqrt{14^{2}-4\left(-5\right)\times 3}}{2\left(-5\right)}$$

Square 14.

$$t=\frac{-14±\sqrt{196-4\left(-5\right)\times 3}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-14±\sqrt{196+20\times 3}}{2\left(-5\right)}$$

Multiply 20 times 3.

$$t=\frac{-14±\sqrt{196+60}}{2\left(-5\right)}$$

Add 196 to 60.

$$t=\frac{-14±\sqrt{256}}{2\left(-5\right)}$$

Take the square root of 256.

$$t=\frac{-14±16}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-14±16}{-10}$$

Now solve the equation t=\frac{-14±16}{-10} when ± is plus. Add -14 to 16.

$$t=\frac{2}{-10}$$

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

$$t=-\frac{1}{5}$$

Now solve the equation t=\frac{-14±16}{-10} when ± is minus. Subtract 16 from -14.

$$t=-\frac{30}{-10}$$

Divide -30 by -10.

$$t=3$$

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

$$-5t^{2}+14t+3=-5\left(t-\left(-\frac{1}{5}\right)\right)\left(t-3\right)$$

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

$$-5t^{2}+14t+3=-5\left(t+\frac{1}{5}\right)\left(t-3\right)$$

Add \frac{1}{5} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-5t^{2}+14t+3=-5\times \frac{-5t-1}{-5}\left(t-3\right)$$

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

$$-5t^{2}+14t+3=\left(-5t-1\right)\left(t-3\right)$$

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.

$$x ^ 2 -\frac{14}{5}x -\frac{3}{5} = 0$$

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

$$r + s = \frac{14}{5} rs = -\frac{3}{5}$$

Two numbers r and s sum up to \frac{14}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{14}{5} = \frac{7}{5}. 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>

$$r = \frac{7}{5} - u s = \frac{7}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{5}

$$(\frac{7}{5} - u) (\frac{7}{5} + u) = -\frac{3}{5}$$

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

$$\frac{49}{25} - u^2 = -\frac{3}{5}$$

Simplify the expression by subtracting \frac{49}{25} on both sides

$$-u^2 = -\frac{3}{5}-\frac{49}{25} = -\frac{64}{25}$$

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

$$u^2 = \frac{64}{25} u = \pm\sqrt{\frac{64}{25}} = \pm \frac{8}{5}$$

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

$$r =\frac{7}{5} - \frac{8}{5} = -0.200 s = \frac{7}{5} + \frac{8}{5} = 3$$

Evaluate

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