Solve 5t^2+12t+7

$$5 t ^ { 2 } + 12 t + 7$$

Factor

$$\left(t+1\right)\left(5t+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5t^{2}+at+bt+7. To find a and b, set up a system to be solved.

$$a+b=12 ab=5\times 7=35$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 12.

$$a=5 b=7$$

Rewrite 5t^{2}+12t+7 as \left(5t^{2}+5t\right)+\left(7t+7\right).

$$\left(5t^{2}+5t\right)+\left(7t+7\right)$$

Factor out 5t in the first and 7 in the second group.

$$5t\left(t+1\right)+7\left(t+1\right)$$

Factor out common term t+1 by using distributive property.

$$\left(t+1\right)\left(5t+7\right)$$

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.

$$5t^{2}+12t+7=0$$

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.

$$t=\frac{-12±\sqrt{12^{2}-4\times 5\times 7}}{2\times 5}$$

Square 12.

$$t=\frac{-12±\sqrt{144-4\times 5\times 7}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-12±\sqrt{144-20\times 7}}{2\times 5}$$

Multiply -20 times 7.

$$t=\frac{-12±\sqrt{144-140}}{2\times 5}$$

Add 144 to -140.

$$t=\frac{-12±\sqrt{4}}{2\times 5}$$

Take the square root of 4.

$$t=\frac{-12±2}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-12±2}{10}$$

Now solve the equation t=\frac{-12±2}{10} when ± is plus. Add -12 to 2.

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

Divide -10 by 10.

$$t=-1$$

Now solve the equation t=\frac{-12±2}{10} when ± is minus. Subtract 2 from -12.

$$t=-\frac{14}{10}$$

Reduce the fraction \frac{-14}{10} to lowest terms by extracting and canceling out 2.

$$t=-\frac{7}{5}$$

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{7}{5} for x_{2}.

$$5t^{2}+12t+7=5\left(t-\left(-1\right)\right)\left(t-\left(-\frac{7}{5}\right)\right)$$

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

$$5t^{2}+12t+7=5\left(t+1\right)\left(t+\frac{7}{5}\right)$$

Add \frac{7}{5} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5t^{2}+12t+7=5\left(t+1\right)\times \frac{5t+7}{5}$$

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

$$5t^{2}+12t+7=\left(t+1\right)\left(5t+7\right)$$

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 5

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

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

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

Two numbers r and s sum up to -\frac{12}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{12}{5} = -\frac{6}{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>

$$r = -\frac{6}{5} - u s = -\frac{6}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{5}

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

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

$$\frac{36}{25} - u^2 = \frac{7}{5}$$

Simplify the expression by subtracting \frac{36}{25} on both sides

$$-u^2 = \frac{7}{5}-\frac{36}{25} = -\frac{1}{25}$$

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

$$u^2 = \frac{1}{25} u = \pm\sqrt{\frac{1}{25}} = \pm \frac{1}{5}$$

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

$$r =-\frac{6}{5} - \frac{1}{5} = -1.400 s = -\frac{6}{5} + \frac{1}{5} = -1.000$$

Evaluate

$$\left(t+1\right)\left(5t+7\right)$$