Solve 5x^2+3x-18

$$5 x ^ { 2 } + 3 x - 18$$

Factor

$$5\left(x-\frac{-3\sqrt{41}-3}{10}\right)\left(x-\frac{3\sqrt{41}-3}{10}\right)$$

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.

$$5x^{2}+3x-18=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-18\right)}}{2\times 5}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 5\left(-18\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-3±\sqrt{9-20\left(-18\right)}}{2\times 5}$$

Multiply -20 times -18.

$$x=\frac{-3±\sqrt{9+360}}{2\times 5}$$

Add 9 to 360.

$$x=\frac{-3±\sqrt{369}}{2\times 5}$$

Take the square root of 369.

$$x=\frac{-3±3\sqrt{41}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-3±3\sqrt{41}}{10}$$

Now solve the equation x=\frac{-3±3\sqrt{41}}{10} when ± is plus. Add -3 to 3\sqrt{41}.

$$x=\frac{3\sqrt{41}-3}{10}$$

Now solve the equation x=\frac{-3±3\sqrt{41}}{10} when ± is minus. Subtract 3\sqrt{41} from -3.

$$x=\frac{-3\sqrt{41}-3}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+3\sqrt{41}}{10} for x_{1} and \frac{-3-3\sqrt{41}}{10} for x_{2}.

$$5x^{2}+3x-18=5\left(x-\frac{3\sqrt{41}-3}{10}\right)\left(x-\frac{-3\sqrt{41}-3}{10}\right)$$

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

$$x ^ 2 +\frac{3}{5}x -\frac{18}{5} = 0$$

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

$$r + s = -\frac{3}{5} rs = -\frac{18}{5}$$

Two numbers r and s sum up to -\frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{5} = -\frac{3}{10}. 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>

$$r = -\frac{3}{10} - u s = -\frac{3}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{18}{5}

$$(-\frac{3}{10} - u) (-\frac{3}{10} + u) = -\frac{18}{5}$$

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

$$\frac{9}{100} - u^2 = -\frac{18}{5}$$

Simplify the expression by subtracting \frac{9}{100} on both sides

$$-u^2 = -\frac{18}{5}-\frac{9}{100} = -\frac{369}{100}$$

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

$$u^2 = \frac{369}{100} u = \pm\sqrt{\frac{369}{100}} = \pm \frac{\sqrt{369}}{10}$$

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

$$r =-\frac{3}{10} - \frac{\sqrt{369}}{10} = -2.221 s = -\frac{3}{10} + \frac{\sqrt{369}}{10} = 1.621$$

Evaluate

$$5x^{2}+3x-18$$