Question 396016
1.) 4x^2+x-5=0

I will solve one with detailed steps. Rest i am sure you can walk through


Find the roots of the equation by quadratic formula
a= 4  , b = 1  , c =  -5
b^2-4ac= 1*1 - (4*4*(-5))= 81  .

{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}
{{{x1= (-(1)+sqrt(81))/(2*4)}}}
{{{x1=(-1+sqrt(81))/8}}}
x1=1
...
{{{x2=(-b-sqrt(b^2-4ac))/(2a)}}}
{{{x2=(-1-sqrt(81))/8}}}
x2= -1.25

1, -1.25 are the roots of the equation.



2.) 2x^2-5x+2=0
a=2,b=-5,c=2

3.)3x^2+7=-6x

4.) 7x^2-x-12=0