Solve 7x^2+52x+67

$$7 x ^ { 2 } + 52 x + 67$$

Factor

$$7\left(x-\frac{-3\sqrt{23}-26}{7}\right)\left(x-\frac{3\sqrt{23}-26}{7}\right)$$

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.

$$7x^{2}+52x+67=0$$

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.

$$x=\frac{-52±\sqrt{52^{2}-4\times 7\times 67}}{2\times 7}$$

Square 52.

$$x=\frac{-52±\sqrt{2704-4\times 7\times 67}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-52±\sqrt{2704-28\times 67}}{2\times 7}$$

Multiply -28 times 67.

$$x=\frac{-52±\sqrt{2704-1876}}{2\times 7}$$

Add 2704 to -1876.

$$x=\frac{-52±\sqrt{828}}{2\times 7}$$

Take the square root of 828.

$$x=\frac{-52±6\sqrt{23}}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-52±6\sqrt{23}}{14}$$

Now solve the equation x=\frac{-52±6\sqrt{23}}{14} when ± is plus. Add -52 to 6\sqrt{23}.

$$x=\frac{6\sqrt{23}-52}{14}$$

Divide -52+6\sqrt{23} by 14.

$$x=\frac{3\sqrt{23}-26}{7}$$

Now solve the equation x=\frac{-52±6\sqrt{23}}{14} when ± is minus. Subtract 6\sqrt{23} from -52.

$$x=\frac{-6\sqrt{23}-52}{14}$$

Divide -52-6\sqrt{23} by 14.

$$x=\frac{-3\sqrt{23}-26}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-26+3\sqrt{23}}{7} for x_{1} and \frac{-26-3\sqrt{23}}{7} for x_{2}.

$$7x^{2}+52x+67=7\left(x-\frac{3\sqrt{23}-26}{7}\right)\left(x-\frac{-3\sqrt{23}-26}{7}\right)$$

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 7

$$x ^ 2 +\frac{52}{7}x +\frac{67}{7} = 0$$

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

$$r + s = -\frac{52}{7} rs = \frac{67}{7}$$

Two numbers r and s sum up to -\frac{52}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{52}{7} = -\frac{26}{7}. 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>

$$r = -\frac{26}{7} - u s = -\frac{26}{7} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{67}{7}

$$(-\frac{26}{7} - u) (-\frac{26}{7} + u) = \frac{67}{7}$$

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

$$\frac{676}{49} - u^2 = \frac{67}{7}$$

Simplify the expression by subtracting \frac{676}{49} on both sides

$$-u^2 = \frac{67}{7}-\frac{676}{49} = -\frac{207}{49}$$

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

$$u^2 = \frac{207}{49} u = \pm\sqrt{\frac{207}{49}} = \pm \frac{\sqrt{207}}{7}$$

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

$$r =-\frac{26}{7} - \frac{\sqrt{207}}{7} = -5.770 s = -\frac{26}{7} + \frac{\sqrt{207}}{7} = -1.659$$

Evaluate

$$7x^{2}+52x+67$$