# SOLUTION: 42. Find b^2 - 4ac and the number of real solutions to each equation. 14. Solve each equation by using the quadratic formula. -8q^2+2q+1=0

Algebra ->  Algebra  -> Radicals -> SOLUTION: 42. Find b^2 - 4ac and the number of real solutions to each equation. 14. Solve each equation by using the quadratic formula. -8q^2+2q+1=0      Log On

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 Algebra: Radicals -- complicated equations involving roots Solvers Lessons Answers archive Quiz In Depth

 Question 255667: 42. Find b^2 - 4ac and the number of real solutions to each equation. 14. Solve each equation by using the quadratic formula. -8q^2+2q+1=0Answer by stanbon(57347)   (Show Source): You can put this solution on YOUR website!42. Find b^2 - 4ac and the number of real solutions to each equation. If you have a quadrate: y = ax^2 + bx + c The roots are x = [-b +- sqrt(b^2-4ac)]/(2a) --- If b^2 - 4ac = 0 the two roots are both x = -b/2a --- If b^2 - 4ac > 0 you will have two Real Number solutions --- If b^2 -4ac < 0 you will get two Complex Number solutions ----------------------------------------------------------------- 14. Solve each equation by using the quadratic formula. -8q^2+2q+1=0 x = [-2 +- sqrt(4 - 4*-8*1)]/(-16) --- x = [-2 +- sqrt(36)]/(-16) --- x = [-2 +- 6]/(-16) --- x = [-8/-16] or x = [4/-16] x = 1/2 or x = -1/4 ======================= Cheers, Stan H.