How do you solve $5x ^ { 2} + 5x - 1= 0$ ?

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Answer 1 · 2017-03-12T08:11:51

$x=0.171$ or $x=-1.171$

$5x^2+5x-1=0$

Standard form of a quadratic equation:

$ax^2+bx+c=0$

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

$:.x=((-5)+-sqrt(5^2-(4*5*-1)))/(2*5)$

$:.x=(-5+-sqrt(25+20))/10$

$:.x=(-5+sqrt45)/10$ or $x=(-5-sqrt45)/10$

$:.x=(-5+6.708203933)/10$ or $x=(-5-6.708203933)/10$

$:.x=1.708203933/10$ or $x=-11.708203933/10$

$:.x=0.1708203933$ or $x=-1.1708203933$

$:.x=0.171$ or $x=-1.171$

How do you solve 5x ^ { 2} + 5x - 1= 05x ^ { 2} + 5x - 1= 0?