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2x^2+3x-5=0

2x^2+3x-5=0 equation

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Numerical solution:

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The solution

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2*x  + 3*x - 5 = 0
$$\left(2 x^{2} + 3 x\right) - 5 = 0$$
Detail solution
This equation is of the form
a*x^2 + b*x + c = 0

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
$$a = 2$$
$$b = 3$$
$$c = -5$$
, then
D = b^2 - 4 * a * c = 

(3)^2 - 4 * (2) * (-5) = 49

Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = 1$$
$$x_{2} = - \frac{5}{2}$$
Vieta's Theorem
rewrite the equation
$$\left(2 x^{2} + 3 x\right) - 5 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{3 x}{2} - \frac{5}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{3}{2}$$
$$q = \frac{c}{a}$$
$$q = - \frac{5}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{3}{2}$$
$$x_{1} x_{2} = - \frac{5}{2}$$
The graph
Sum and product of roots [src]
sum
1 - 5/2
$$- \frac{5}{2} + 1$$
=
-3/2
$$- \frac{3}{2}$$
product
-5/2
$$- \frac{5}{2}$$
=
-5/2
$$- \frac{5}{2}$$
-5/2
Rapid solution [src]
x1 = -5/2
$$x_{1} = - \frac{5}{2}$$
x2 = 1
$$x_{2} = 1$$
x2 = 1
Numerical answer [src]
x1 = 1.0
x2 = -2.5
x2 = -2.5
The graph
2x^2+3x-5=0 equation