Respuesta :

danielmaduroh

Answer:

[tex]x_{1}=1 \ and \ x_{2}=-\frac{7}{4}[/tex]

Step-by-step explanation:

The equation:

[tex]-4x^2-3x+7=0[/tex]

Can be solved by using the quadratic formula:

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ where \\ a=-4 \\ b=-3 \\ c=7 \\ \\ x=\frac{-(-3)\pm\sqrt{(-3)^2-4(-4)(7)}}{2(-4)} \\ \\ x=\frac{3\pm\sqrt{9+112}}{-8} \\ \\ where: \\ \\ x_{1}=1 \ and \ x_{2}=-\frac{7}{4}[/tex]

These two values represents the zeroes of the polynomial function [tex]f(x)=-4x^2-3x+7[/tex]

zainebamir540

Answer:

Quadratic equation and  two distinct real solutions

Step-by-step explanation:

Given equation is

-4x²-3x+7 = 0

We have to determine the type of equation and number of solution.

For type of equation,

Highest power of equation is 2.

Hence, equation is quadratic equation.

Now,For number of solution:

ax²+bx+c = 0 is general quadratic equation.

Comparing general equation with given equation, we have

a = -4, b = -3 and  c = 7

We use the formula of discriminant.

D = b²-4ac

D = (-3)²-4(-4)(7)

D = 9+112

D = 121 > 0

Hence, Discriminant is real.

Hence, there are two distinct  real solutions of given equation.