How do you solve $5x^2-6x-4=0$ using the quadratic formula?

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Answer 1 · 2018-04-13T11:35:00

The roots are $(3+sqrt29)/5$ or $(3-sqrt29)/5$ .

$x=(-b±sqrt(b^2-4ac))/(2a)$

$x=6±sqrt((-6)^2-4*5*(-4))/(2*5)$

$x=(6±sqrt(36+80))/(10)$

$x=(6±sqrt(116))/(10)$

$x=(6±2sqrt(29))/(10)$

$x=(3±sqrt(29))/(5)$

Therefore the roots can be $(3+sqrt29)/5$ or $(3-sqrt29)/5$ .

Answer 2 · 2018-04-13T11:44:05

Solution: $x= 3/5 +- sqrt 29 /5$

$5 x^2- 6 x -4 =0$

Comparing with standard quadratic equation $ax^2+bx+c=0$

$a= 5 ,b=-6 ,c=- 4$ . Discriminant, $D= b^2-4 a c$ or

$D=36+80 =116$ , discriminant is positive, we get two real

solutions, Quadratic formula: $x= (-b+-sqrtD)/(2a)$ or

$x= (6 +- sqrt 116)/10 = 3/5 +- sqrt 29 /5$

Solution: $x= 3/5 +- sqrt 29 /5$ [Ans]

How do you solve 5x^2-6x-4=05x^2-6x-4=0 using the quadratic formula?