13 mark for review which quadratic equation has no real solutions? a (x^{2}+14x - 49 = 0) b (x^{2}-14x + 49…

Explanation:

Step1: Recall discriminant formula

For a quadratic equation $ax^{2}+bx + c = 0$, the discriminant $\Delta=b^{2}-4ac$. If $\Delta<0$, there are no real - solutions.

Step2: Analyze option A

For $x^{2}+14x - 49 = 0$, where $a = 1$, $b = 14$, $c=-49$. Then $\Delta=(14)^{2}-4\times1\times(-49)=196 + 196=392>0$.

Step3: Analyze option B

For $x^{2}-14x + 49 = 0$, where $a = 1$, $b=-14$, $c = 49$. Then $\Delta=(-14)^{2}-4\times1\times49=196 - 196=0$.

Step4: Analyze option C

For $5x^{2}-14x - 49 = 0$, where $a = 5$, $b=-14$, $c=-49$. Then $\Delta=(-14)^{2}-4\times5\times(-49)=196+980 = 1176>0$.

Step5: Analyze option D

For $5x^{2}-14x + 49 = 0$, where $a = 5$, $b=-14$, $c = 49$. Then $\Delta=(-14)^{2}-4\times5\times49=196 - 980=-784<0$.

Answer:

D. $5x^{2}-14x + 49 = 0$