Solve =-7x^2+15x-5

$$= - 7 x ^ { 2 } + 15 x - 5$$

Factor

$$-7\left(x-\frac{15-\sqrt{85}}{14}\right)\left(x-\frac{\sqrt{85}+15}{14}\right)$$

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.

$$-7x^{2}+15x-5=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\left(-7\right)\left(-5\right)}}{2\left(-7\right)}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\left(-7\right)\left(-5\right)}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-15±\sqrt{225+28\left(-5\right)}}{2\left(-7\right)}$$

Multiply 28 times -5.

$$x=\frac{-15±\sqrt{225-140}}{2\left(-7\right)}$$

Add 225 to -140.

$$x=\frac{-15±\sqrt{85}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-15±\sqrt{85}}{-14}$$

Now solve the equation x=\frac{-15±\sqrt{85}}{-14} when ± is plus. Add -15 to \sqrt{85}.

$$x=\frac{\sqrt{85}-15}{-14}$$

Divide -15+\sqrt{85} by -14.

$$x=\frac{15-\sqrt{85}}{14}$$

Now solve the equation x=\frac{-15±\sqrt{85}}{-14} when ± is minus. Subtract \sqrt{85} from -15.

$$x=\frac{-\sqrt{85}-15}{-14}$$

Divide -15-\sqrt{85} by -14.

$$x=\frac{\sqrt{85}+15}{14}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{15-\sqrt{85}}{14} for x_{1} and \frac{15+\sqrt{85}}{14} for x_{2}.

$$-7x^{2}+15x-5=-7\left(x-\frac{15-\sqrt{85}}{14}\right)\left(x-\frac{\sqrt{85}+15}{14}\right)$$

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.

$$x ^ 2 -\frac{15}{7}x +\frac{5}{7} = 0$$

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

$$r + s = \frac{15}{7} rs = \frac{5}{7}$$

Two numbers r and s sum up to \frac{15}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{15}{7} = \frac{15}{14}. 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>

$$r = \frac{15}{14} - u s = \frac{15}{14} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{7}

$$(\frac{15}{14} - u) (\frac{15}{14} + u) = \frac{5}{7}$$

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

$$\frac{225}{196} - u^2 = \frac{5}{7}$$

Simplify the expression by subtracting \frac{225}{196} on both sides

$$-u^2 = \frac{5}{7}-\frac{225}{196} = -\frac{85}{196}$$

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

$$u^2 = \frac{85}{196} u = \pm\sqrt{\frac{85}{196}} = \pm \frac{\sqrt{85}}{14}$$

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

$$r =\frac{15}{14} - \frac{\sqrt{85}}{14} = 0.413 s = \frac{15}{14} + \frac{\sqrt{85}}{14} = 1.730$$

Evaluate

$$-7x^{2}+15x-5$$