Question

solve through formula method:
x^2+3x-28=0​

Answer 1

Solution:

x²+3x-28=0 ...(1)

We need to solve it using quadratic formula which says if a quadratic equation is

ax²+bx+c=0 ...(2)

then its discriminant is given by

D = b² - 4ac

then its roots are

x = ( -b ± √D )/2a

Comparing equations (1) and (2) we get

a = 1 , b = 3 , c = - 28

Now finding its Discriminant :-

D = b² - 4ac

D = 3² - 4×1×-28

D = 9 + 112

D = 121

Now finding roots of the given equation by quadratic formula :-

x = ( -3 + √121 )/ 2×1

x = ( -3 + 11 )/ 2

x = 4

And

x = ( -3 - √121 )/ 2×1

x = ( -3 - 11)/ 2×1

x = -7

Hence, roots of the given equation are 4 and -7.

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• $\sf{Solve :\:x^2+3x-28=0}$

★═════════════════★  

♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{x=4,\:x=-7}}$

★═════════════════★

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\underline{\underline{\sf{x^2+3x-28=0}}}$

Solve with the quadratic formula

$\bf{Quadratic\:Equation\:Formula:}$

$\sf{\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}}$

$\sf{x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}$

For

• a = 1

• b = 3

• c = -28

$\sf{x_{1,\:2}=\dfrac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \left(-28\right)}}{2\cdot \:1}}$

$\sf{x_{1,\:2}=\dfrac{-3\pm \sqrt{3^2+4\cdot \:1\cdot \:28}}{2\cdot 1}}$

$\sf{x_{1,\:2}=\dfrac{-3\pm \sqrt{3^2+112}}{2\cdot 1}}$

$\sf{x_{1,\:2}=\dfrac{-3\pm \sqrt{9+112}}{2\cdot 1}}$

$\sf{x_{1,\:2}=\dfrac{-3\pm \sqrt{121}}{2\cdot 1}}$

$\sf{x_{1,\:2}=\dfrac{-3\pm \sqrt{11^2}}{2\cdot 1}}$

$\sf{x_{1,\:2}=\dfrac{-3\pm \:11}{2\cdot \:1}}$

Separate the solutions

$\sf{\displaystyle x_1=\frac{-3+11}{2\cdot \:1},\:x_2=\frac{-3-11}{2\cdot \:1}}$

$--------------------$

$\sf{x=\dfrac{-3+11}{2\cdot \:1}}$

$\sf{x=\dfrac{8}{2\cdot \:1}}$

$\sf{x=\dfrac{8}{2}}$

$\bf{x=4}$

$--------------------$

$\sf{\displaystyle x=\frac{-3-11}{2\cdot \:1}}$

$\sf{x=\dfrac{-14}{2\cdot \:1}}$

$\sf{x=\dfrac{-14}{2}}$

$\bf{x=-7}$

$--------------------$

The solutions to the quadratic equation are :

$\large\boxed{\sf{x=4,\:x=-7}}$