Solve 5u^2+30u-70

$$5 u ^ { 2 } + 30 u - 70$$

Factor

$$5\left(u-\left(-\sqrt{23}-3\right)\right)\left(u-\left(\sqrt{23}-3\right)\right)$$

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.

$$5u^{2}+30u-70=0$$

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.

$$u=\frac{-30±\sqrt{30^{2}-4\times 5\left(-70\right)}}{2\times 5}$$

Square 30.

$$u=\frac{-30±\sqrt{900-4\times 5\left(-70\right)}}{2\times 5}$$

Multiply -4 times 5.

$$u=\frac{-30±\sqrt{900-20\left(-70\right)}}{2\times 5}$$

Multiply -20 times -70.

$$u=\frac{-30±\sqrt{900+1400}}{2\times 5}$$

Add 900 to 1400.

$$u=\frac{-30±\sqrt{2300}}{2\times 5}$$

Take the square root of 2300.

$$u=\frac{-30±10\sqrt{23}}{2\times 5}$$

Multiply 2 times 5.

$$u=\frac{-30±10\sqrt{23}}{10}$$

Now solve the equation u=\frac{-30±10\sqrt{23}}{10} when ± is plus. Add -30 to 10\sqrt{23}.

$$u=\frac{10\sqrt{23}-30}{10}$$

Divide -30+10\sqrt{23} by 10.

$$u=\sqrt{23}-3$$

Now solve the equation u=\frac{-30±10\sqrt{23}}{10} when ± is minus. Subtract 10\sqrt{23} from -30.

$$u=\frac{-10\sqrt{23}-30}{10}$$

Divide -30-10\sqrt{23} by 10.

$$u=-\sqrt{23}-3$$

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

$$5u^{2}+30u-70=5\left(u-\left(\sqrt{23}-3\right)\right)\left(u-\left(-\sqrt{23}-3\right)\right)$$

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

$$x ^ 2 +6x -14 = 0$$

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

$$r + s = -6 rs = -14$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -14

$$(-3 - u) (-3 + u) = -14$$

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

$$9 - u^2 = -14$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = -14-9 = -23$$

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

$$u^2 = 23 u = \pm\sqrt{23} = \pm \sqrt{23}$$

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

$$r =-3 - \sqrt{23} = -7.796 s = -3 + \sqrt{23} = 1.796$$

Evaluate

$$5\left(u^{2}+6u-14\right)$$