A quadratic equation can be solved using the discriminant. The roots of the quadratic equation: $$x_{1} = \frac{\sqrt{D} - b}{2 a}$$ $$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$ where D = b^2 - 4*a*c - it is the discriminant. Because $$a = 2$$ $$b = -5$$ $$c = -3$$ , then
D = b^2 - 4 * a * c =
(-5)^2 - 4 * (2) * (-3) = 49
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or $$x_{1} = 3$$ $$x_{2} = - \frac{1}{2}$$
Vieta's Theorem
rewrite the equation $$\left(2 x^{2} - 5 x\right) - 3 = 0$$ of $$a x^{2} + b x + c = 0$$ as reduced quadratic equation $$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$ $$x^{2} - \frac{5 x}{2} - \frac{3}{2} = 0$$ $$p x + q + x^{2} = 0$$ where $$p = \frac{b}{a}$$ $$p = - \frac{5}{2}$$ $$q = \frac{c}{a}$$ $$q = - \frac{3}{2}$$ Vieta Formulas $$x_{1} + x_{2} = - p$$ $$x_{1} x_{2} = q$$ $$x_{1} + x_{2} = \frac{5}{2}$$ $$x_{1} x_{2} = - \frac{3}{2}$$