Solve -16t^2+6t-28

$$- 16 t ^ { 2 } + 6 t - 28$$

Factor

$$2\left(-8t^{2}+3t-14\right)$$

Solution Steps

Factor out 2. Polynomial -8t^{2}+3t-14 is not factored since it does not have any rational roots.

$$2\left(-8t^{2}+3t-14\right)$$

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.

$$-16t^{2}+6t-28=0$$

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.

$$t=\frac{-6±\sqrt{6^{2}-4\left(-16\right)\left(-28\right)}}{2\left(-16\right)}$$

Square 6.

$$t=\frac{-6±\sqrt{36-4\left(-16\right)\left(-28\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-6±\sqrt{36+64\left(-28\right)}}{2\left(-16\right)}$$

Multiply 64 times -28.

$$t=\frac{-6±\sqrt{36-1792}}{2\left(-16\right)}$$

Add 36 to -1792.

$$t=\frac{-6±\sqrt{-1756}}{2\left(-16\right)}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$-16t^{2}+6t-28$$

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.

$$x ^ 2 -\frac{3}{8}x +\frac{7}{4} = 0$$

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

$$r + s = \frac{3}{8} rs = \frac{7}{4}$$

Two numbers r and s sum up to \frac{3}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{8} = \frac{3}{16}. 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>

$$r = \frac{3}{16} - u s = \frac{3}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{4}

$$(\frac{3}{16} - u) (\frac{3}{16} + u) = \frac{7}{4}$$

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

$$\frac{9}{256} - u^2 = \frac{7}{4}$$

Simplify the expression by subtracting \frac{9}{256} on both sides

$$-u^2 = \frac{7}{4}-\frac{9}{256} = \frac{439}{256}$$

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

$$u^2 = -\frac{439}{256} u = \pm\sqrt{-\frac{439}{256}} = \pm \frac{\sqrt{439}}{16}i$$

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

$$r =\frac{3}{16} - \frac{\sqrt{439}}{16}i = 0.188 - 1.310i s = \frac{3}{16} + \frac{\sqrt{439}}{16}i = 0.188 + 1.310i$$

Evaluate

$$-16t^{2}+6t-28$$