Solve -40a^2+19a+30

$$- 40 a ^ { 2 } + 19 a + 30$$

Factor

$$-40\left(a-\frac{19-\sqrt{5161}}{80}\right)\left(a-\frac{\sqrt{5161}+19}{80}\right)$$

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.

$$-40a^{2}+19a+30=0$$

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.

$$a=\frac{-19±\sqrt{19^{2}-4\left(-40\right)\times 30}}{2\left(-40\right)}$$

Square 19.

$$a=\frac{-19±\sqrt{361-4\left(-40\right)\times 30}}{2\left(-40\right)}$$

Multiply -4 times -40.

$$a=\frac{-19±\sqrt{361+160\times 30}}{2\left(-40\right)}$$

Multiply 160 times 30.

$$a=\frac{-19±\sqrt{361+4800}}{2\left(-40\right)}$$

Add 361 to 4800.

$$a=\frac{-19±\sqrt{5161}}{2\left(-40\right)}$$

Multiply 2 times -40.

$$a=\frac{-19±\sqrt{5161}}{-80}$$

Now solve the equation a=\frac{-19±\sqrt{5161}}{-80} when ± is plus. Add -19 to \sqrt{5161}.

$$a=\frac{\sqrt{5161}-19}{-80}$$

Divide -19+\sqrt{5161} by -80.

$$a=\frac{19-\sqrt{5161}}{80}$$

Now solve the equation a=\frac{-19±\sqrt{5161}}{-80} when ± is minus. Subtract \sqrt{5161} from -19.

$$a=\frac{-\sqrt{5161}-19}{-80}$$

Divide -19-\sqrt{5161} by -80.

$$a=\frac{\sqrt{5161}+19}{80}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{19-\sqrt{5161}}{80} for x_{1} and \frac{19+\sqrt{5161}}{80} for x_{2}.

$$-40a^{2}+19a+30=-40\left(a-\frac{19-\sqrt{5161}}{80}\right)\left(a-\frac{\sqrt{5161}+19}{80}\right)$$

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.

$$x ^ 2 -\frac{19}{40}x -\frac{3}{4} = 0$$

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

$$r + s = \frac{19}{40} rs = -\frac{3}{4}$$

Two numbers r and s sum up to \frac{19}{40} exactly when the average of the two numbers is \frac{1}{2}*\frac{19}{40} = \frac{19}{80}. 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>

$$r = \frac{19}{80} - u s = \frac{19}{80} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}

$$(\frac{19}{80} - u) (\frac{19}{80} + u) = -\frac{3}{4}$$

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

$$\frac{361}{6400} - u^2 = -\frac{3}{4}$$

Simplify the expression by subtracting \frac{361}{6400} on both sides

$$-u^2 = -\frac{3}{4}-\frac{361}{6400} = -\frac{5161}{6400}$$

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

$$u^2 = \frac{5161}{6400} u = \pm\sqrt{\frac{5161}{6400}} = \pm \frac{\sqrt{5161}}{80}$$

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

$$r =\frac{19}{80} - \frac{\sqrt{5161}}{80} = -0.661 s = \frac{19}{80} + \frac{\sqrt{5161}}{80} = 1.136$$

Evaluate

$$30+19a-40a^{2}$$