Solve f(1)=6x^2+5x-7

$$f ( 1 ) = 6 x ^ { 2 } + 5 x - 7$$

Factor

$$6\left(x-\frac{-\sqrt{193}-5}{12}\right)\left(x-\frac{\sqrt{193}-5}{12}\right)$$

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.

$$6x^{2}+5x-7=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 6\left(-7\right)}}{2\times 6}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 6\left(-7\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-5±\sqrt{25-24\left(-7\right)}}{2\times 6}$$

Multiply -24 times -7.

$$x=\frac{-5±\sqrt{25+168}}{2\times 6}$$

Add 25 to 168.

$$x=\frac{-5±\sqrt{193}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-5±\sqrt{193}}{12}$$

Now solve the equation x=\frac{-5±\sqrt{193}}{12} when ± is plus. Add -5 to \sqrt{193}.

$$x=\frac{\sqrt{193}-5}{12}$$

Now solve the equation x=\frac{-5±\sqrt{193}}{12} when ± is minus. Subtract \sqrt{193} from -5.

$$x=\frac{-\sqrt{193}-5}{12}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{193}}{12} for x_{1} and \frac{-5-\sqrt{193}}{12} for x_{2}.

$$6x^{2}+5x-7=6\left(x-\frac{\sqrt{193}-5}{12}\right)\left(x-\frac{-\sqrt{193}-5}{12}\right)$$

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 6

$$x ^ 2 +\frac{5}{6}x -\frac{7}{6} = 0$$

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

$$r + s = -\frac{5}{6} rs = -\frac{7}{6}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{6}

$$(-\frac{5}{12} - u) (-\frac{5}{12} + u) = -\frac{7}{6}$$

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

$$\frac{25}{144} - u^2 = -\frac{7}{6}$$

Simplify the expression by subtracting \frac{25}{144} on both sides

$$-u^2 = -\frac{7}{6}-\frac{25}{144} = -\frac{193}{144}$$

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

$$u^2 = \frac{193}{144} u = \pm\sqrt{\frac{193}{144}} = \pm \frac{\sqrt{193}}{12}$$

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

$$r =-\frac{5}{12} - \frac{\sqrt{193}}{12} = -1.574 s = -\frac{5}{12} + \frac{\sqrt{193}}{12} = 0.741$$

Evaluate

$$6x^{2}+5x-7$$