9/20Find the number of real number solutions for the equation. x2 + 5x + 7 = 0nd01ations02O cannot be determinedare00e100%

Question:Find the number of real number solutions for the equation. x^2 + 5x + 7 = 0 Answer:The number of real solutions for the equation [tex]x^{2}+5 x+7=0[/tex] is zeroSolution:For a Quadratic Equation of form : [tex]a x^{2}+b x+c=0[/tex]  ---- eqn 1The solution is [tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]  Now , the given Quadratic Equation is [tex]x^{2}+5 x+7=0[/tex]  ---- eqn 2On comparing Equation (1) and Equation(2), we geta = 1 , b = 5 and c = 7 In [tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex] , [tex]b^2 - 4ac[/tex] is called the discriminant of the quadratic equationIts value determines the nature of roots Now, here are the rules with discriminants: 1) D > 0; there are 2 real solutions in the equation 2) D = 0; there is 1 real solution in the equation 3) D < 0; there are no real solutions in the equationNow let solve for given equation[tex]D= b^2 - 4ac\\\\D = 5^2 - 4(1)(7)\\\\D = 25 - 28 \\\\D = -3[/tex]Since -3 is less than 0, this means that there are 0 real solutions in this equation.