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4b^2+17b-21=0 equation

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Numerical solution:

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The solution

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4*b  + 17*b - 21 = 0
$$\left(4 b^{2} + 17 b\right) - 21 = 0$$
Detail solution
This equation is of the form
a*b^2 + b*b + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$b_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$b_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 4$$
$$b = 17$$
$$c = -21$$
, then
D = b^2 - 4 * a * c = 

(17)^2 - 4 * (4) * (-21) = 625

Because D > 0, then the equation has two roots.
b1 = (-b + sqrt(D)) / (2*a)

b2 = (-b - sqrt(D)) / (2*a)

or
$$b_{1} = 1$$
$$b_{2} = - \frac{21}{4}$$
Vieta's Theorem
rewrite the equation
$$\left(4 b^{2} + 17 b\right) - 21 = 0$$
of
$$a b^{2} + b^{2} + c = 0$$
as reduced quadratic equation
$$b^{2} + \frac{b^{2}}{a} + \frac{c}{a} = 0$$
$$b^{2} + \frac{17 b}{4} - \frac{21}{4} = 0$$
$$b^{2} + b p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{17}{4}$$
$$q = \frac{c}{a}$$
$$q = - \frac{21}{4}$$
Vieta Formulas
$$b_{1} + b_{2} = - p$$
$$b_{1} b_{2} = q$$
$$b_{1} + b_{2} = - \frac{17}{4}$$
$$b_{1} b_{2} = - \frac{21}{4}$$
The graph
Sum and product of roots [src]
sum
1 - 21/4
$$- \frac{21}{4} + 1$$
=
-17/4
$$- \frac{17}{4}$$
product
-21/4
$$- \frac{21}{4}$$
=
-21/4
$$- \frac{21}{4}$$
-21/4
Rapid solution [src]
b1 = -21/4
$$b_{1} = - \frac{21}{4}$$
b2 = 1
$$b_{2} = 1$$
b2 = 1
Numerical answer [src]
b1 = 1.0
b2 = -5.25
b2 = -5.25