Answer:
[tex]\large\boxed{x=\dfrac{-9\pm\sqrt{41}}{2}}[/tex]
Step-by-step explanation:
[tex]\text{The quadratic formula of}[/tex]
[tex]ax^2+bx+c=0[/tex]
[tex]\text{If}\ b^2-4a<0,\ \text{then the equation has no solution}\\\\\text{If}\ b^2-4ac=0,\ \text{then the equation has one solution}\ x=\dfrac{-b}{2a}\\\\\text{If}\ b^2-4ac>0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\=========================================[/tex]
[tex]\text{We have}\ x^2+9x+10=0\\\\a=1,\ b=9,\ c=10\\\\\text{substitute:}\\\\b^2-4ac=9^2-4(1)(10)=81-40=41>0\qquad _{\text{two solutions}}\\\\\sqrt{b^2-4ac}=\sqrt{41}\\\\x=\dfrac{-9\pm\sqrt{41}}{2(1)}=\dfrac{-9\pm\sqrt{41}}{2}[/tex]
Answer:
Step-by-step explanation:
Points to remember
solution of a quadratic equation ax² + bx + c = 0
x = [-b ± √(b² - 4ac)]/2a
It is given that, x² + 9x + 10 = 0.
To find the solution
Here a = 1, b= 9 and c = 10
x = [-b ± √(b² - 4ac)]/2a
= [-9 ± √(9² - 4*1*10)]/2*1
= [-9 ± √(81 - 40)]/2
= [-9 ± √(41)]/2
= [-9 ± √41]/2
Therefore the solution of given quadratic equation = [-9 ± √41]/2
