Solve alpha^2-0,8alpha+3,125beta=0

$$\alpha ^ { 2 } - 0,8 \alpha + 3,125 \beta = 0$$

Solve for β

$$\beta =-\frac{8\alpha \left(\alpha -0,8\right)}{25}$$

Steps for Solving Linear Equation

Subtract \alpha ^{2} from both sides. Anything subtracted from zero gives its negation.

$$-0,8\alpha +3,125\beta =-\alpha ^{2}$$

Add 0,8\alpha to both sides.

$$3,125\beta =-\alpha ^{2}+0,8\alpha$$

The equation is in standard form.

$$3,125\beta =-\alpha ^{2}+\frac{4\alpha }{5}$$

Divide both sides of the equation by 3,125, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{3,125\beta }{3,125}=\frac{\alpha \left(0,8-\alpha \right)}{3,125}$$

Dividing by 3,125 undoes the multiplication by 3,125.

$$\beta =\frac{\alpha \left(0,8-\alpha \right)}{3,125}$$

Divide \alpha \left(0,8-\alpha \right) by 3,125 by multiplying \alpha \left(0,8-\alpha \right) by the reciprocal of 3,125.

$$\beta =\frac{8\alpha \left(0,8-\alpha \right)}{25}$$

Solve for α

$$\alpha =\frac{\sqrt{-\frac{25\beta }{2}+0,64}}{2}+0,4$$

$$\alpha =-\frac{\sqrt{-\frac{25\beta }{2}+0,64}}{2}+0,4\text{, }\beta \leq 0,0512$$