# How do you solve  2c^2 - 7c = -5?

Apr 5, 2016

1 and 5/2

#### Explanation:

$y = 2 {c}^{2} - 7 c + 5 = 0$.
Since a + b + c = 0, use shortcut. The 2 real rots are: 1 and $\frac{c}{a} = \frac{5}{2}$

Apr 5, 2016

$x = \frac{5}{2} , 1$

#### Explanation:

color(blue)(2c^2-7c=-5

Add $5$ both sides

$\rightarrow 2 {c}^{2} - 7 c + 5 = - 5 + 5$

$\rightarrow 2 {c}^{2} - 7 c + 5 = 0$

Now,this is a trinomial.

So,

We can solve this by factoring or using Quadratic equation

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Factoring

color(purple)(2c^2-7c+5=0

Factor the equation

If you have any problem with factoring trinomials,Watch this video:

$\rightarrow \left(2 x - 5\right) \left(x - 1\right) = 0$

Now we can say that

1)color(orange)((x-1)=0

2)color(indigo)((2x-5)=0

Solve for both of the equations

1)color(orange)(x-1=0

color(green)(rArrx=1

2)color(indigo)(2x-5=0

$\rightarrow 2 x = 5$

color(green)(rArrx=5/2

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

color(purple)(2c^2-7+5=0

This is a Quadratic equation (in form $a {x}^{2} + b x + c = 0$)

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Remember that $a , b \mathmr{and} c$ are the coefficients

So,

color(violet)(a=2,b=-7,c=5

$\rightarrow x = \frac{- \left(- 7\right) \pm \sqrt{- {7}^{2} - 4 \left(2\right) \left(5\right)}}{2 \left(2\right)}$

$\rightarrow x = \frac{7 \pm \sqrt{49 - 4 \left(40\right)}}{4}$

$\rightarrow x = \frac{7 \pm \sqrt{49 - 40}}{4}$

$\rightarrow x = \frac{7 \pm \sqrt{9}}{4}$

$\rightarrow x = \frac{7 \pm 3}{4}$

Now we have two solutions

1)color(orange)(x=(7+3)/(4)=10/4=5/2

2)color(indigo)(x=(7-3)/4=4/4=1

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

:.color(blue)(ul bar |x=5/2,1|