Verified answer

x1, x2 = [-b ± √(b² - 4ac)] / 2a

[11 ± √((-11)² - 4(2)(10))] / (2(2))

[11 ± √41] / 4

Answer: [11 ± √41] / 4

Do you know the quadratic formula ? Do you know the standard form for the quadratic equation ? You need to read your book if you don't. When you have found these concepts, you will be able to substitute the right values in the quadratic formula.

x = (-b ± √(b^2 - 4ac))/2a is the quadratic formula. It will be in your book.

we are able to apply this rule for the two equations (-b +/- sqrt(b^2-4ac)) /2a a) x^2+7x+2=0 a =a , b=7 , c=2 (-7 +/- sqrt(40 9-8)) / 2 = (-7 +/- sqrt(40-one)) / 2 = (-7 +/- 6.4) /2 =((-7+6.4)/2) , ((-7-6.4)/2) x= - 0.3 or x= -6.7 --------------------------------------... b)2x^2+11x-15=0 a=2 , b=11, c=-15 (-11 +/- sqrt(121-4*2*-15))/ 2*2 = (-11 +/- sqrt( 121+one hundred twenty)) / 4 = (-11 +/- sqrt(241)) / 4 = (-11 +/- 15.fifty two) /4 = ((-11+15.fifty two) / 4) , ((-11-15.fifty two) / 4) x= a million.13 or x= -6.sixty 3

Deltaa=121-80=41,x1={+11+(41) ^ 1/2}/4 & x2={+11-(41) ^1/2}/4