Question 61750
Solve the following quadratic equations by using quadratic formula. 
2.1y^2 - 3.5y = 4
: 
2.1y^2 - 3.5y - 4 = 0
I assume you know that the quadratic formula is, here is the first one step-by=step:
 a = 2.1; b = -3.5; c = -4
:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{y = (-(-3.5) +- sqrt( (-3.5^2)-4*2.1*-4 ))/(2*2.1) }}}
{{{y = (+3.5 +- sqrt(12.25 - (-33.6)  ))/(4.2) }}}
{{{y = (+3.5 +- sqrt(12.25 + 33.6 ))/(4.2) }}}
{{{y = (+3.5 +- sqrt(45.85 ))/(4.2) }}}
{{{y = (+3.5 + 6.77)/(4.2) }}}
{{{y = 10.27/4.2 }}}
y = +2.446
and
{{{y = (+3.5 - 6.77)/(4.2) }}}
{{{y = -3.27/4.2 }}}
y = -.779
:
:
a(a+2) = -1
a^2 + 2a + 1 = 0; 
a = 1; b = 2; c = 1,
{{{a = (-2 +- sqrt( 2^2-4*1*1 ))/(2*1) }}}
Complete this like we did above, you will find that it is easy, everything inside the radical cancels out. 
a = -1
:
:
Substitute for a,b,c like we did in the 1st one
6x^2 + 5x -4 = 3x - 2
6x^2 + 5x - 3x - 4 + 2 = 0
6x^2 + 2x - 2 = 0
a = 6; b = 2; c = -2
:
: 
4x^2 - 14x - 27 = 3
4x^2 - 14x - 27 - 3 = 0
4x^2 - 14x - 30 = 0
a = 4; b =-14; c = -30