Solve 2x^2+42x-54

$$2 x ^ { 2 } + 42 x - 54$$

Factor

$$2\left(x-\frac{-3\sqrt{61}-21}{2}\right)\left(x-\frac{3\sqrt{61}-21}{2}\right)$$

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.

$$2x^{2}+42x-54=0$$

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.

$$x=\frac{-42±\sqrt{42^{2}-4\times 2\left(-54\right)}}{2\times 2}$$

Square 42.

$$x=\frac{-42±\sqrt{1764-4\times 2\left(-54\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-42±\sqrt{1764-8\left(-54\right)}}{2\times 2}$$

Multiply -8 times -54.

$$x=\frac{-42±\sqrt{1764+432}}{2\times 2}$$

Add 1764 to 432.

$$x=\frac{-42±\sqrt{2196}}{2\times 2}$$

Take the square root of 2196.

$$x=\frac{-42±6\sqrt{61}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-42±6\sqrt{61}}{4}$$

Now solve the equation x=\frac{-42±6\sqrt{61}}{4} when ± is plus. Add -42 to 6\sqrt{61}.

$$x=\frac{6\sqrt{61}-42}{4}$$

Divide -42+6\sqrt{61} by 4.

$$x=\frac{3\sqrt{61}-21}{2}$$

Now solve the equation x=\frac{-42±6\sqrt{61}}{4} when ± is minus. Subtract 6\sqrt{61} from -42.

$$x=\frac{-6\sqrt{61}-42}{4}$$

Divide -42-6\sqrt{61} by 4.

$$x=\frac{-3\sqrt{61}-21}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-21+3\sqrt{61}}{2} for x_{1} and \frac{-21-3\sqrt{61}}{2} for x_{2}.

$$2x^{2}+42x-54=2\left(x-\frac{3\sqrt{61}-21}{2}\right)\left(x-\frac{-3\sqrt{61}-21}{2}\right)$$

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 2

$$x ^ 2 +21x -27 = 0$$

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

$$r + s = -21 rs = -27$$

Two numbers r and s sum up to -21 exactly when the average of the two numbers is \frac{1}{2}*-21 = -\frac{21}{2}. 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>

$$r = -\frac{21}{2} - u s = -\frac{21}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -27

$$(-\frac{21}{2} - u) (-\frac{21}{2} + u) = -27$$

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

$$\frac{441}{4} - u^2 = -27$$

Simplify the expression by subtracting \frac{441}{4} on both sides

$$-u^2 = -27-\frac{441}{4} = -\frac{549}{4}$$

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

$$u^2 = \frac{549}{4} u = \pm\sqrt{\frac{549}{4}} = \pm \frac{\sqrt{549}}{2}$$

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

$$r =-\frac{21}{2} - \frac{\sqrt{549}}{2} = -22.215 s = -\frac{21}{2} + \frac{\sqrt{549}}{2} = 1.215$$

Evaluate

$$2\left(x^{2}+21x-27\right)$$